Category: Capacitance

Q: A capacitor is charged by a battery . The battery is removed and another identical uncharged capacitor is connected in parallel . The total electrostatic energy of resulting system :

(a) Decreases by a factor 2

(b) Remains same

(c) Increases by a factor 2

(d) Increases by a factor 4

Ans: (a)

Sol: $\large U_i = \frac{1}{2}CV^2$

Common Potential $V’ = \frac{C_1V_1 + C_2 V_2}{C_1+C_2}$

$V’ = \frac{CV + 0}{C+C}$

$V’ = \frac{V}{2}$

$\large U_f = \frac{1}{2}(C+C)(\frac{V}{2})^2$

$\large U_f = \frac{1}{4}CV^2 $

$\large \frac{U_f}{U_i} = \frac{1}{2}$

Q: A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity S and mass m. If the temperature of the block is raised by  ΔT, the potential difference V across the capacitor is

(a) $ \displaystyle \sqrt{\frac{2 m C \Delta T}{S} }$

(b) $ \displaystyle \frac{ m C \Delta T}{S} $

(c) $ \displaystyle \frac{ m S \Delta T}{C} $

(d) $ \displaystyle \sqrt{\frac{2 m S \Delta T}{C}} $

Ans: (d)

Sol:

$ \displaystyle \frac{1}{2}C V^2 = m S \Delta T$

$ \displaystyle V = \sqrt{\frac{2 m S \Delta T}{C}} $

Q: In the given figure the capacitor of plate area A is charged upto charge q. The ratio of elongations (neglect force of gravity) in springs C and D at equilibrium position is

(a) $ \displaystyle \frac{k_1}{k_2} $

(b) $ \displaystyle \frac{k_2}{k_1} $

(c) $ \displaystyle k_1 k_2 $

(d) $ \displaystyle \sqrt{ \frac{k_1}{k_2}} $

Ans: (b)

Sol:

Force acting between plates of capacitor is attractive & it is

$ \displaystyle F = \frac{q^2}{2 \epsilon_0 A} $

$\displaystyle k_1 x_1 = k_2 x_2 $

$\displaystyle \frac{x_1}{x_2} = \frac{k_2}{k_1} $

Q: In figure, given C₁ = 3 μF, C₂ = 5 μF, C₃ = 9 μF, and C₄ = 13 μF . What is the potential difference between points A and B ?

(a) 13 V

(b) 9 V

(c) 0 V

(d) 11 V

Ans: (a)

Sol: V1 + V2 = 22 …(i)

9V1 = 13V2 …(ii)

Q: Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be (charge on each condenser is q₀; k = dielectric constant)

(a) $ \displaystyle \frac{2 q_0}{1+\frac{1}{k}}$

(b) $ \displaystyle \frac{ q_0}{1+\frac{1}{k}}$

(c) $ \displaystyle \frac{2 q_0}{1+k}$

(d) $ \displaystyle \frac{ q_0}{1+k}$

Ans: (a)

Sol:
$ q_0/C + q_0/C = 2 $ …(i)

After introduction of slab , let charge be q₁

$q_1/C + q_1/kC = 2$ …(ii)