Ncert Solutions for Class 12 Physics Chapter 2 – Electrostatic Potential And Capacitance

Question 2.1:

Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Answer:

There are two charges,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2c42eba8

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges, as shown in the given figure.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html mb348d22

r = Distance of point P from charge q1

Let the electric potential (V) at point P be zero.

Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m31ec219

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9= Permittivity of free space

For V = 0, equation (i) reduces to

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7a331e9d

Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

Suppose point P is outside the system of two charges at a distance from the negative charge, where potential is zero, as shown in the following figure.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m6ea7d84b

For this arrangement, potential is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 514f421e

For V = 0, equation (ii) reduces to

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m4869ba04

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

Question 2.2:

A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer:

The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m535fb361

Where,

Charge, q = 5 µC = 5 × 10−6 C

Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm

Distance of each vertex from centre O, d = 10 cm

Electric potential at point O,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m1576e6be

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9= Permittivity of free space

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m2e2ae644

Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.

Question 2.3:

Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.

(a) Identify an equipotential surface of the system.

(b) What is the direction of the electric field at every point on this surface?

Answer:

(a) The situation is represented in the given figure.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m6851d289

An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.

(b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.

Question 2.4:

A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field

(a) Inside the sphere

(b) Just outside the sphere

(c) At a point 18 cm from the centre of the sphere?

Answer:

(a) Radius of the spherical conductor, r = 12 cm = 0.12 m

Charge is uniformly distributed over the conductor, = 1.6 × 10−7 C

Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.

(b) Electric field E just outside the conductor is given by the relation,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6da84e69

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9= Permittivity of free space

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 89bbc47

Therefore, the electric field just outside the sphere is NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 69571664.

(c) Electric field at a point 18 m from the centre of the sphere = E1

Distance of the point from the centre, d = 18 cm = 0.18 m

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 4d2d48e6

Therefore, the electric field at a point 18 cm from the centre of the sphere is

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2caad74f.

Question 2.5:

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Answer:

Capacitance between the parallel plates of the capacitor, C = 8 pF

Initially, distance between the parallel plates was and it was filled with air. Dielectric constant of air, k = 1

Capacitance, C, is given by the formula,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m48d1918b

Where,

A = Area of each plate

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9= Permittivity of free space

If distance between the plates is reduced to half, then new distance, d = NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m83168b7

Dielectric constant of the substance filled in between the plates, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2ffc626 = 6

Hence, capacitance of the capacitor becomes

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m75b35eac

Taking ratios of equations (i) and (ii), we obtain

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m61660651

Therefore, the capacitance between the plates is 96 pF.

Question 2.6:

Three capacitors each of capacitance 9 pF are connected in series.

(a) What is the total capacitance of the combination?

(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Answer:

(a) Capacitance of each of the three capacitors, C = 9 pF

Equivalent capacitance (C) of the combination of the capacitors is given by the relation,

1C’=1C+1C+1C⇒1C’=19+19+19=13⇒C’=3 pFTherefore, total capacitance of the combination is

3 pF.

(b) Supply voltage, V = 120 V

Potential difference (V‘) across each capacitor is equal to one-third of the supply voltage.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 10b7257e

Therefore, the potential difference across each capacitor is 40 V.

Question 2.7:

Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

(a) What is the total capacitance of the combination?

(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Answer:

(a) Capacitances of the given capacitors are

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6f32fb45

For the parallel combination of the capacitors, equivalent capacitorNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 27919028is given by the algebraic sum,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m11932738

Therefore, total capacitance of the combination is 9 pF.

(b) Supply voltage, V = 100 V

The voltage through all the three capacitors is same = V = 100 V

Charge on a capacitor of capacitance C and potential difference V is given by the relation,

q = VC … (i)

For C = 2 pF,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m37ac10cc

For C = 3 pF,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m3f5cccb8

For C = 4 pF,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2ebd8a67

Question 2.8:

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Answer:

Area of each plate of the parallel plate capacitor, A = 6 × 10−3 m2

Distance between the plates, d = 3 mm = 3 × 10−3 m

Supply voltage, V = 100 V

Capacitance C of a parallel plate capacitor is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2d913553

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9= Permittivity of free space

= 8.854 × 10−12 N−1 m−2 C−2

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 386ef4e3

Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 × 10−9 C.

Page No 88:

Question 2.9:

Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

(a) While the voltage supply remained connected.

(b) After the supply was disconnected.

Answer:

(a) Dielectric constant of the mica sheet, k = 6

Initial capacitance, C = 1.771 × 10−11 F

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 5f39de4c

Supply voltage, = 100 V

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m728d88a5

Potential across the plates remains 100 V.

(b) Dielectric constant, k = 6

Initial capacitance, C = 1.771 × 10−11 F

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 5f39de4c

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge = 1.771 × 10−9 C

Potential across the plates is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 52a8eb26

Question 2.10:

A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Answer:

Capacitor of the capacitance, C = 12 pF = 12 × 10−12 F

Potential difference, = 50 V

Electrostatic energy stored in the capacitor is given by the relation,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 312972ad

Therefore, the electrostatic energy stored in the capacitor is NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html md4089e6

Question 2.11:

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Answer:

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6a61be56

If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C) of the combination is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m57dab833

New electrostatic energy can be calculated as

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m41785325

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 5f5dc8bb

Therefore, the electrostatic energy lost in the process isNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 3d2346d2.

Question 2.12:

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10−9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Answer:

Charge located at the origin, q = 8 mC= 8 × 10−3 C

Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = − 2 × 10−9 C

All the points are represented in the given figure.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m510a9b77

Point P is at a distance, d1 = 3 cm, from the origin along z-axis.

Point Q is at a distance, d2 = 4 cm, from the origin along y-axis.

Potential at point P, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 11cda99d

Potential at point Q, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html e672915

Work done (W) by the electrostatic force is independent of the path.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m40425fce

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 35e77b48

Therefore, work done during the process is 1.27 J.

Question 2.13:

A cube of side has a charge at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Answer:

Length of the side of a cube = b

Charge at each of its vertices = q

A cube of side is shown in the following figure.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 58fd4cf5

d = Diagonal of one of the six faces of the cube

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html mb6388d9

l = Length of the diagonal of the cube

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 574cabce

The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m2f7013cd

Therefore, the potential at the centre of the cube is NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 455403c8.

The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the centre of the cube. Hence, the electric field is zero at the centre.

Question 2.14:

Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

(a) at the mid-point of the line joining the two charges, and

(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Answer:

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 3f3ff29b

Magnitude of charge located at A, q1 = 1.5 μC

Magnitude of charge located at B, q2 = 2.5 μC

Distance between the two charges, d = 30 cm = 0.3 m

(a) Let V1 and E1 are the electric potential and electric field respectively at O.

V1 = Potential due to charge at A + Potential due to charge at B

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7d4639fe

Where,

0 = Permittivity of free space

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m5b082891

E1 = Electric field due to q2 − Electric field due to q1

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 77a10c06

Therefore, the potential at mid-point is 2.4 × 105 V and the electric field at mid-point is 4× 105 V m−1. The field is directed from the larger charge to the smaller charge.

(b) Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in the following figure.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m26363935

V2 and Eare the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6108b1ea

V2= Electric potential due to A + Electric Potential due to B

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 1ff1255e

Electric field due to q at Z,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m7c18cb7

Electric field due to q2 at Z,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m279bff7c

The resultant field intensity at Z,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html aee0753

Where, 2θis the angle, ∠AZ B

From the figure, we obtain

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 59c9b80e

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 105 V and electric field is 6.6 ×105 V m−1.

Question 2.15:

A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

(a) A charge is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Answer:

(a) Charge placed at the centre of a shell is +q. Hence, a charge of magnitude −q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is −q.

Surface charge density at the inner surface of the shell is given by the relation,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m1f9e9d79

A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m53c7daa8

(b) Yes

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

Question 2.16:

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 3af4bf33

Where

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 21845f56is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 21845f56is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 60e8cad2

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Answer:

(a) Electric field on one side of a charged body is E1 and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 4ebcb0d

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 21845f56 = Unit vector normal to the surface at a point

σ = Surface charge density at that point

Electric field due to the other surface of the charged body,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m1b416077

Electric field at any point due to the two surfaces,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m2d48eeaa

Since inside a closed conductor, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 15801455 = 0,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m2c7a56a2

Therefore, the electric field just outside the conductor is NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 68f895e4.

(b) When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

Question 2.17:

A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Answer:

Charge density of the long charged cylinder of length L and radius r is λ.

Another cylinder of same length surrounds the pervious cylinder. The radius of this cylinder is R.

Let E be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gauss’s theorem as,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 41bad03a

Where, = Distance of a point from the common axis of the cylinders

Let q be the total charge on the cylinder.

It can be written as

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 796369e0

Where,

q = Charge on the inner sphere of the outer cylinder

0 = Permittivity of free space

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 3035393d

Therefore, the electric field in the space between the two cylinders isNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 24321b.

Question 2.18:

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

Answer:

The distance between electron-proton of a hydrogen atom, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m45448199

Charge on an electron, q1 = −1.6 ×10−19 C

Charge on a proton, q2 = +1.6 ×10−19 C

(a) Potential at infinity is zero.

Potential energy of the system, = Potential energy at infinity − Potential energy at distance d

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m68950c0a

where,

0 is the permittivity of free space

14πε0=9×109 Nm2C-2∴ Potential energy=0-9×109×1.6×10-1920.53×10-10=-43.47×10-19 J∵1.6×10-19 J=1 eV∴Potential energy=-43.7×10-19=-43.7×10-191.6×10-19=-27.2 eV

Therefore, the potential energy of the system is −27.2 eV.

(b) Kinetic energy is half of the magnitude of potential energy.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7e6717a6

Total energy = 13.6 − 27.2 = 13.6 eV

Therefore, the minimum work required to free the electron is 13.6 eV.

(c) When zero of potential energy is taken, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 5a9633b3

∴Potential energy of the system = Potential energy at d1 − Potential energy at d

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html f453b7d

Page No 89:

Question 2.19:

If one of the two electrons of a Hmolecule is removed, we get a hydrogen molecular ionNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m35b54733. In the ground state of anNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m35b54733, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Answer:

The system of two protons and one electron is represented in the given figure.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 34503647

Charge on proton 1, q1 = 1.6 ×10−19 C

Charge on proton 2, q2 = 1.6 ×10−19 C

Charge on electron, q3 = −1.6 ×10−19 C

Distance between protons 1 and 2, d1 = 1.5 ×10−10 m

Distance between proton 1 and electron, d2 = 1 ×10−10 m

Distance between proton 2 and electron, d3 = 1 × 10−10 m

The potential energy at infinity is zero.

Potential energy of the system,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m7f1ceeae

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 19033f6a

Therefore, the potential energy of the system is −19.2 eV.

Question 2.20:

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Answer:

Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.

Let EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m303d5e64

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m6c5e67c6

Putting the value of (2) in (1), we obtain

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m263fcc25

Therefore, the ratio of electric fields at the surface isNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m1e38d54f.

A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.

Question 2.21:

Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points?

(b) Obtain the dependence of potential on the distance of a point from the origin when r/>> 1.

(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Answer:

(a) Zero at both the points

Charge − q is located at (0, 0, − a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (xy, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (xy, 0) is zero. Electrostatic potential at point (0, 0, z) is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 1005577a

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9 = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

(b) Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to square of the distance i.e., NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 48264244

(c) Zero

The answer does not change if the path of the test is not along the x-axis.

A test charge is moved from point (5, 0, 0) to point (−7, 0, 0) along the x-axis. Electrostatic potential (V1) at point (5, 0, 0) is given by,

V1=-q4π∈01(5-0)2+(-a)2+ q4π∈01(5-0)2+(a)2     =-q4π∈025+a2+q4π∈025+a2      =0Electrostatic potential, V2, at point (− 7, 0, 0) is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 28c73175

Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the x-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

Question 2.22:

Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on for r/>> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m81681dd

Answer:

Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 74443307

A point is located at P, which is r distance away from point Y.

The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges.

Charge +placed at point X

Charge −2q placed at point Y

Charge +q placed at point Z

XY = YZ = a

YP = r

PX = r + a

PZ = r − a

Electrostatic potential caused by the system of three charges at point P is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m990dae7

SinceNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 770006cc,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m7c9bef43

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 18602596is taken as negligible.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 39572dd

It can be inferred that potential, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 4c133537

However, it is known that for a dipole, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 48264244

And, for a monopole, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 1453e995

Question 2.23:

An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Answer:

Total required capacitance, C = 2 µF

Potential difference, V = 1 kV = 1000 V

Capacitance of each capacitor, C1 = 1µF

Each capacitor can withstand a potential difference, V1 = 400 V

Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 4f93715b

Hence, there are three capacitors in each row.

Capacitance of each row

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m710b40ed

Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 1b15f406

Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required for the given arrangement.

Question 2.24:

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Answer:

Capacitance of a parallel capacitor, V = 2 F

Distance between the two plates, d = 0.5 cm = 0.5 × 10−2 m

Capacitance of a parallel plate capacitor is given by the relation,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m61741ca2

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9 = Permittivity of free space = 8.85 × 10−12 C2 N−1 m−2

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 771bf961

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.

Page No 90:

Question 2.25:

Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html mc04e1e6

Answer:

Capacitance of capacitor Cis 100 pF.

Capacitance of capacitor Cis 200 pF.

Capacitance of capacitor Cis 200 pF.

Capacitance of capacitor Cis 100 pF.

Supply potential, V = 300 V

Capacitors C2 and C3 are connected in series. Let their equivalent capacitance be NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6e17271

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m1c24e1ea

Capacitors C1 and C’ are in parallel. Let their equivalent capacitance be NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m750723b5

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 5c19e90d

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2ae79860are connected in series. Let their equivalent capacitance be C.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m3886cfae

Hence, the equivalent capacitance of the circuit is NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m717cd02a

Potential difference across NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m58a85617NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 561228c4

Potential difference across C4 = V4

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m6062f6a

Charge on NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m61c9bda

Q4CV

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m22640c95

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 1b90c16d

Hence, potential difference, V1, across C1 is 100 V.

Charge on C1 is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m55a7166a

C2 and C3 having same capacitances have a potential difference of 100 V together. Since C2 and C3 are in series, the potential difference across Cand Cis given by,

V2 = V3 = 50 V

Therefore, charge on C2 is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m322b5bd2

And charge on C3 ­is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m4aa31669

Hence, the equivalent capacitance of the given circuit isNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 555be8a3 NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m4225286e

Question 2.26:

The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between and the magnitude of electric field between the plates.

Answer:

Area of the plates of a parallel plate capacitor, A = 90 cm= 90 × 10−4 m2

Distance between the plates, d = 2.5 mm = 2.5 × 10−3 m

Potential difference across the plates, V = 400 V

(a) Capacitance of the capacitor is given by the relation,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2d913553

Electrostatic energy stored in the capacitor is given by the relation, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7a4f5e5d

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 36bd2a85

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9 = Permittivity of free space = 8.85 × 10−12 C2 N−1 m−2

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 138f11c8

Hence, the electrostatic energy stored by the capacitor is NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 24606061

(b) Volume of the given capacitor,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m1a9e7aaf

Energy stored in the capacitor per unit volume is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6a0ddbfd

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 3bb90975= Electric intensity = E

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6b82521f

Question 2.27:

A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer:

Capacitance of a charged capacitor, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 24ca0f37

Supply voltage, V1 = 200 V

Electrostatic energy stored in Cis given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m47a0baac

Capacitance of an uncharged capacitor, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6b371ad3

When C2 is connected to the circuit, the potential acquired by it is V2.

According to the conservation of charge, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2.

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2355fae1

Electrostatic energy for the combination of two capacitors is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7e665d7f

Hence, amount of electrostatic energy lost by capacitor C1

E1 − E2

= 0.08 − 0.0533 = 0.0267

= 2.67 × 10−2 J

Question 2.28:

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where is the charge on the capacitor, and is the magnitude of electric field between the plates. Explain the origin of the factor ½.

Answer:

Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx

As a result, the potential energy of the capacitor increases by an amount given as uAΔx.

Where,

u = Energy density

A = Area of each plate

d = Distance between the plates

V = Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e.,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 82c109f

Electric intensity is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m89f4ee7

However, capacitance, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 2d913553

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6dfdcc94

Charge on the capacitor is given by,

Q = CV

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 22d45894

The physical origin of the factor, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m5a4d85ce, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 9c31ab, of the field that contributes to the force.

Question 2.29:

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 23d0cf65

that the capacitance of a spherical capacitor is given by

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7e17d091

where r1 and r2 are the radii of outer and inner spheres, respectively.

Answer:

Radius of the outer shell = r1

Radius of the inner shell = r2

The inner surface of the outer shell has charge +Q.

The outer surface of the inner shell has induced charge −Q.

Potential difference between the two shells is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 58de6c54

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9 = Permittivity of free space

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7e77f154

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m45fef145

Hence, proved.

Page No 91:

Question 2.30:

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

Answer:

Radius of the inner sphere, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m5be939b1 = 12 cm = 0.12 m

Radius of the outer sphere, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 8c6bec3= 13 cm = 0.13 m

Charge on the inner sphere,NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 54e886de

Dielectric constant of a liquid, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7bde026c

(a) NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m7916b0f4

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 76437b11

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9= Permittivity of free space = NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 4ca656b5

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 1228129

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 3c86783f

Hence, the capacitance of the capacitor is approximately NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 6519b1b4.

(b) Potential of the inner sphere is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 1d34f42a

Hence, the potential of the inner sphere is NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m5d4a6b80.

(c) Radius of an isolated sphere, = 12 × 10−2 m

Capacitance of the sphere is given by the relation,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7f13c82f

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.

Question 2.31:

Answer carefully:

(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q1 Q2/4πNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d92, where is the distance between their centres?

(b) If Coulomb’s law involved 1/r3 dependence (instead of 1/r2), would Gauss’s law be still true?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

(f) What meaning would you give to the capacitance of a single conductor?

(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Answer:

(a) The force between two conducting spheres is not exactly given by the expression, Q1 Q2/4πNS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d92, because there is a non-uniform charge distribution on the spheres.

(b) Gauss’s law will not be true, if Coulomb’s law involved 1/rdependence, instead of1/r2, on r.

(c) Yes,

If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.

(d) Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

(e) No

Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

(f) The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

(g) Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

Question 2.32:

A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Answer:

Length of a co-axial cylinder, l = 15 cm = 0.15 m

Radius of outer cylinder, r1 = 1.5 cm = 0.015 m

Radius of inner cylinder, r2 = 1.4 cm = 0.014 m

Charge on the inner cylinder, q = 3.5 µC = 3.5 × 10−6 C

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 695fa581

Where,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9 = Permittivity of free space = NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m4e3fbc8b

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m35a3337b

Potential difference of the inner cylinder is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 268ce3b0

Question 2.33:

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Answer:

Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m21c7d47a=3

Dielectric strength = 107 V/m

For safety, the field intensity never exceeds 10% of the dielectric strength.

Hence, electric field intensity, E = 10% of 10= 106 V/m

Capacitance of the parallel plate capacitor, C = 50 pF = 50 × 10−12 F

Distance between the plates is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html m222a125e

Where,

A = Area of each plate

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 149b60d9 = Permittivity of free space =NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 1a256176

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 7be53eea

Hence, the area of each plate is about 19 cm2.

Question 2.34:

Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,

(c) a single positive charge at the origin, and

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Answer:

(a) Equidistant planes parallel to the xy plane are the equipotential surfaces.

(b) Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

(c) Concentric spheres centered at the origin are equipotential surfaces.

(d) A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

Page No 92:

Question 2.35:

In a Van de Graaff type generator a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 107 Vm−1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Answer:

Potential difference, V = 15 × 106 V

Dielectric strength of the surrounding gas = 5 × 107 V/m

Electric field intensity, E = Dielectric strength = 5 × 107 V/m

Minimum radius of the spherical shell required for the purpose is given by,

NS 22 10 08 Sravana 12 Physics 2 37 NRJ SS html 10d055e3

Hence, the minimum radius of the spherical shell required is 30 cm.

Question 2.36:

A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.

Answer:

According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge q1 on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2. For positive charge q1, potential difference is always positive.

Question 2.37:

Answer the following:

(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

Answer:

(a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

(b) Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.

(d) During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.


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