Solutions-Class-12-Physics-Chapter-8-Electrostatics-MSBSHSE

(A) Constant, decreases, decreases

(D) C = \(\frac{ε_0A}{d}(\frac{4k}{k+3})\)

(A) 1:1

(A) \(-\frac{Qq}{6πε_0L}\)

(A) 1.78 × 10^-8 C

The equatorial plane of an electric dipole is an equipotential with V = 0. Therefore, the no work is done in moving a charge between two points in the equatorial plane of a dipole.

The capacitance of a spherical capacitor is C = 4πε₀ \((\frac{ab}{b-a})\) where a and b are the radii of the concentric inner and outer conducting shells

Hence, the capacitance decreases if the difference b—a is increased. '

Suppose the parallel-plate capacitor has capacitance C₀, plates of area A and separation d.  Assume the metal sheet introduced has the same area A.

Case (1) : Finite thickness t. Free electrons in the sheet will migrate towards the positive plate of the capacitor. Then, the metal sheet is attracted towards whichever capacitor plate is closest and gets stuck to it, so that its potential is the same as that of that plate. The gap between the capacitor plates is reduced to d— t, so that the capacitance increases.

Case (2) : Negligible thickness. The thin metal sheet divides the gap into two of thicknesses d₁ and d₂ of capacitances C₁ = ε₀A/d₁ and C₂ = ε₀A/d₂ in series.

Their effective capacitance is

C = \(\frac{c_1c_2}{c_1+c_2}=\frac{ε_0A}{d_1+d_2}\) = \(\frac{ε_0A}{d}\) = C₀

i.e., the capacitance remains unchanged.

• There is danger of lightning strikes during a thunderstorm. Because trees are taller than people and therefore closer to the clouds above, they are more likely to get hit by lightning.
• Similarly, a person standing in open ground is the tallest object and more likely to get hit by a lightning.
• But car with a metal body is an almost ideal Faraday cage. When a car is struck by lightning, the charge flows on the outside surface of the car to the ground but the electric field inside remains zero. This leaves the passengers inside unharmed.

Consider a spherical conducting shell of radius r placed in a medium of permittivity s. The mechanical force per unit area on the charged conductor is

F= F/dS = σ²/2ε

where σ is the surface charge density on the conductor. Given the charge on the spherical shell is Q,

σ = Q/4πr². The force acts outward, normal to the surface.

Suppose the force displaces a charged area element σdS through a small distance dx, then the work done by the force is

dW = Fdx = \((\frac{σ^2}{2ε}ds)dx\)

During the displacement, the area element sweeps out a volume dV = dS.dx.

Since V = \(\frac{4}{3}\)πr²  dV = 4πr²dr

∴ dW = \(\frac{σ^2}{2ε}\)dV = \(\frac{1}{2ε}(\frac{Q}{4πr^2})^2(4πr^2dr)\)

         = \(\frac{Q}{8πεr^2}dr\)

Therefore, the work done by the force in expanding the shell from radius r = b to r = a is

W = \(\int dw = \frac{Q}{8πε}\int_b^a \frac{1}{r^2}dr = \frac{Q}{8πε}[-\frac{1}{r}]_b^a\) = \(\frac{Q}{8πε}[\frac{1}{b}-\frac{1}{a}]\)

This gives the required expression for the work done.

(a) Given, the equipotentials of the external uniform electric field are planes parallel to the yz plane, the electric field \(\vec{E} = ±E\hat{i}\) , that is, \(\vec{E}\) is parallel to the x-axis.

(b) From Fig. the dipole moment, \(\vec{E}\) = q(2b)\(\hat{j}\)

The torque on this dipole,

\(\vec{τ}\) = \(\vec{p}\) x \(\vec{E}\)  = 2qb\(\hat{j}\) x (±E\(\hat{i}\) ) = (2qbE)(\(\hat{i}\) x \(\hat{j}\))

Since  \(\hat{i}\) x \(\hat{j}\) = \(-\hat{k}\) ,

= (±2qbE)(\(-\hat{k}\)) = (2qbE)(±\(\hat{k}\))

So that the magnitude of the torque is τ = 2qbE.

If \(\vec{E}\) is in the direction of the + x-axis, the torque \(\vec{τ}\) is in the direction of — z-axis, while if \(\vec{E}\) is in the direction of the —x-axis, the torque \(\vec{τ}\) is in the direction of + z-axis.

In Fig. the line joining the charges is shown as the x-axis with the origin at the + Q charge.

Let q₁ = +Q, and q₂ = q₃ = −q.

Let the two −q charges be at ( −a, 0) and (a, 0), since the charges are given to be equidistant.

r₂₁ = r₃₁ =a and r₃₂ = 2a

The total potential energy of the system of three charges is

U₃ = \(\frac{1}{4πε_0}(\frac{q_1q_2}{r_{21}}+\frac{q_1q_3}{r_{31}}+\frac{q_2q_3}{r_{32}})\)

      = \(\frac{1}{4πε_0}(\frac{(-q)Q}{a}+\frac{Q(-q)}{a}+\frac{(-q)(-q)}{2a})\)

      = \(\frac{1}{4πε_0}(-\frac{2qQ}{a}+\frac{q^2}{2a})\)

Given U₃ = 0

∴ \(\frac{2qQ}{a}=\frac{q^2}{2a}\)

∴ Q/q = ¼

This is the required ratio.

Assume a parallel-plate capacitor, of plate area A and plate separation d is filled with a dielectric of relative permittivity (dielectric constant) k.

Its k capacitance is C = \(\frac{kε_0A}{d}\)  …….(1)

If it is charged to a voltage (potential) V, the charge on its plates is Q = CV.

Since the battery is disconnected after it is charged, the charge Q on its plates, and consequently the product CV, remain unchanged.

On removing the dielectric completely, its capacitance becomes from Eq. (1),

C’ = \(\frac{ε_0A}{d}\) = \(\frac{1}{k}C\)

that is, its capacitance decreases by the factor k-

Since C'V’ = CV, its new voltage is

V’ = \(\frac{C}{C'}V\) = kV

so that its voltage increases by the factor k

The stored potential energy, U =  \(\frac{1}{2}QV\) so that Q remaining constant, U increases by the factor k.

The electric field, E = V/d, so that E also increases by a factor k.

Given : \(\frac{C_1}{C_2}=\frac{1}{2}\) = E, U₁ (for parallel) = U₂ (for series)

\(\frac{C_1}{C_2}=\frac{1}{2}\) ∴ C₂ = 2C₁

For the parallel combination of C₁ and C₂,

C_P = C₁ + C₂ = 3C₁

and charged to a potential V₁, the energy stored is

U₁ = \(\frac{1}{2}\)C_PV₁² = \(\frac{3}{2}\)C₁V₁²

For the series combination of C₁ and C₂,

C_s = \(\frac{C_1C_2}{C_1+C_2}\) = \(\frac{2C_1^2}{3C_1^2}\) = \(\frac{2}{3}C_1\)

and charged to a potential V₂, the energy stored is

U₂ = \(\frac{1}{2}\)C_sV₂² = \(\frac{1}{3}\)C₁V₂²

∴ for U₁ = U₂ = \(\frac{3}{2}\)C₁V₁² = \(\frac{1}{3}\)C₁V₂²

∴ \((\frac{V_1}{V_2})^2=\frac{2}{9}\) ∴ \(\frac{V_1}{V_2}=\frac{\sqrt{2}}{3}=\frac{1.414}{3}\)= 0.471

This gives the required ratio.

The sphere of radius 4a encloses only the negative charge Q₁ = −4Q. The positive charge Q₂ = + 2Q being located at a distance of 5a from the origin is outside the sphere. Only a part of the electric flux lines originating at Q₂ enters the sphere and exits entirely at other points. Hence, the electric flux through the sphere is only due to Q₁.

Therefore, the net electric flux through the sphere = Q₁/ε₀ = −4Q/ε₀. The minus sign shows that the flux is directed into the sphere, but not radially since the sphere is not centred on Q₁.

Given : C = 6 μF = 6 x 10^—6 F = C₁, V = 300 V, C₂ = 3 μF

The electrostatic energy in the capacitor

= \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)(6 x 10^—6)(300)²

= 3 x 10^—6 x 9 x 10⁴ = 0.27 J

The charge on this capacitor,

Q = CV = (6 x 10^—6)(300) = 1.8 mC

When two capacitors of capacitances C₁ and C₂ are connected in parallel, the equivalent capacitance C

=C₁ + C₂ = 6 + 3 = 9 μF

= 9 x 10^—6 F

By conservation of charge, Q = 1.8 C.

∴ The energy of the system = \(\frac{Q^2}{2C}=\frac{(1.8×10^{-3})^2}{2(9×10^{-6})}\) = 0.18 J

The energy lost = 0.27 − 0.18 = 0.09 J = 9 x 10^-2 J

Given: n = 125, q = 0.5 x10⁻⁶ C, d=0.1 m

The radius of each small drop, r = d/2 = 0.05 m

The volume of the larger drop being equal to the volume of the n smaller drops, the radius of the larger drop is

R = \(\sqrt[3]{n}r\) = \(\sqrt[3]{125}(0.05)\) = 0.25 m

The charge on the larger drop,

Q = nq = 125 x (0.5 x 10^—6)C

The electric potential of the surface of the larger drop,

V = \(\frac{1}{4πε_0}\frac{Q}{R}\) = (9 x 10⁹) x \(\frac{125(0.5×10^{-6})}{0.25}\)

    = 9 x 125 x 2 x 10³ = 2.25 x 10⁶ V

Given : p = 6.3 x 10^–30  C-m, N = 10²¹ molecules, E=2.5 x 10⁵ N/C,

θ₀ = θ₁ = 0°, θ = θ₂ = 90°

W= pE(cos θ₀ — cos θ)

The total work required to orient N dipoles is

W = NpE(cos θ₁ — cos θ₂)

= (10²¹)(6.3 x 10^–30)(2.5 x 10⁵)

= 15.75 x 10^–4 J = 1.575 mJ

Given : q₁ = 6 x 10⁻⁶ C, q₂ = −5 x 10⁻⁶ C, A = (0, 6.0 m), P = (8.0m, 0),

r₁ = OP = 8 m, q = e = 1.6 x 10⁻¹⁹ C, 1/4πε₀ = 9 x 10⁹ N-m²/C²

r₂ = AP = \(\sqrt{(8-0)^2+(0-6)^2}\) = \(\sqrt{64+36}\) = 10 m

(a) The net electric potential at P due to the system of two charges is

V = V₁ + V₂ = \(\frac{1}{4πε_0}[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\)

   =  (9 x 10⁹)\([\frac{6×10^{-6}}{8}+\frac{-5×10^{-6}}{10}]\)

   = (9 x 10³)(0.75 − 0.5) = 2.25 x 10³ V

   = 2.25 kV

(b) The electric potential V at the point P is the negative of the work done per unit charge, by the electric field of the system of the charges q₁ and q₂, in bringing a test charge from infinity to that point.

V = −W/q₀

∴ W = −qV = −(1.6 x 10⁻¹⁹)(2.25 x 10³)

= −3.6 x 10⁻¹⁶ J = −2.25 keV

That is, in bringing the positively charged proton from a point of lower potential to a point of higher potential, the work done by the electric field on it is negative, which means that an external agent must bring the proton against the electric field of the system of the two source charges.

Given : k =1 (air), A = 6 × 10^–3 m², d = 2 mm = 2 × 10^–3 m, V = 100 V,

t = 2 mm =d, k₁ = 6, ε₀ = 8.85 x 10^–12 F/m

(a) The capacitance of the air capacitor,

C₀ = \(\frac{ε_0A}{d}\) = \(\frac{(8.85×10^{-12})(6×10^{-3})}{2×10^{-3}}\)

= 26.55 x 10^–12 F = 26.55 pF

(b) Q₀ = C₀V = (26.55 x 10^–12)(100)

= 26.55 x 10^–10 C = 2.655 nC

(c) The dielectric of relative permittivity k₁ completely fills the space between the plates (‘.’ t = d), so that the new capacitance is C =k₁C₀.

With the supply still connected, V remains the same.

∴ Q = CV = kC₀V = kQ₀ = 6(2.655 nF) = 15.93 nC

Therefore, the charge on the plates increases.

[Note : C = k₁C₀= 6 (26.55 pF) = 159.3 pF.]

(i) The capacitor in Fig. (a) is a series combination of three capacitors of plate separations d/3 and plate areas A, with C₁ filled with air (k₁ = 1), C₂ filled with dielectric of k₂ =3 and C₃ filled with dielectric of k₃ = 6.

∴ C₁ = \(\frac{k_1ε_0A}{d/3}=\frac{3ε_0A}{d}k_1\), C₂ = \(\frac{k_2ε_0A}{d/3}=\frac{3ε_0A}{d}k_2\), C₃ = \(\frac{k_3ε_0A}{d/3}=\frac{3ε_0A}{d}k_3\)

∴ \(\frac{1}{C'}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\) = \(\frac{d}{3ε_0A}(\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3})\)

∴ C’ = \(\frac{d}{3ε_0A}(\frac{k_1k_2k_3}{k_1k_2+k_2k_3+k_3k_1})\) = \(\frac{d}{3ε_0A}(\frac{18}{27}\)

       = \(\frac{2ε_0A}{d}\)

(ii) In Fig. (b), a series combination of two capacitors C₂(k₂ =3) and C₃ (k₃ =6), of plate areas A/2 and plate separations d/2, is in parallel with a capacitor C₁ (k₁ =4) of plate area A/2 and plate separation d.

∴ C₁ = \(\frac{k_1ε_0(A/2)}{d}=\frac{ε_0A}{2d}k_1\), C₂ = \(\frac{k_2ε_0(A/2)}{d/2}=\frac{ε_0A}{d}k_2\), C₃ = \(\frac{k_3ε_0(A/2)}{d/2}=\frac{ε_0A}{d}k_3\)

For the series combination of C₂ and C₃,

\(\frac{1}{C_4}=\frac{1}{C_2}+\frac{1}{C_3}\)

∴ C₄ = \(\frac{C_2C_3}{C_2+C_3} = \frac{ε_0A}{d}(\frac{k_2k_3}{k_2+k_3})\)

         = \(\frac{ε_0A}{d}(\frac{3×6}{3+6})\) = \(\frac{2ε_0A}{d}\)

Finally, for the parallel combination of C₁ and C₄,

C” = C₁ + C₄ = \(\frac{ε_0A}{2d}(4)+\frac{2ε_0A}{d}\) = \(\frac{4ε_0A}{d}\)

Thus, the equivalent capacitances are C’ = \(\frac{2ε_0A}{d}\) and C” = \(\frac{4ε_0A}{d}\).