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The work done in placing a charge of 8 × 10^–18 coulomb on a condenser of capacity 100 micro-farad is

32× 10^–32Joule

16 × 10^–32Joule

3.1 × 10^–26Joule

4 × 10^–10Joule

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Explanation

$W = \frac{Q^{2}}{2 C} = \frac{{(8 \times 10^{- 18})}^{2}}{2 \times 100 \times 10^{- 6}} = 32 \times 10^{- 32} J$

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A parallel plate capacitor of capacity C₀ is charged to a potential V₀

(i) The energy stored in the capacitor when the battery is disconnected and the separation is doubled E₁

(ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E₂. Then E₁ / E₂ value is

3/2

1/2

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Explanation

Let $E = \frac{1}{2} C_{0} {V_{0}}^{2} then E_{1} = 2 E$ and $E_{2} = \frac{E}{2}$

So $\frac{E_{1}}{E_{2}} = \frac{4}{1}$

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Two identical capacitors are joined in parallel, charged to a potential V and then separated and then connected in series i.e. the positive plate of one is connected to negative of the other

The charges on the free plates connected together are destroyed

The charges on the free plates are enhanced

The energy stored in the system increases

The potential difference in the free plates becomes 2V

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Explanation

Q₁ = CV and Q₂ = CV

Applying charge conservation $C V_{1} + C V_{2} = Q_{1} + Q_{2}$

$C V_{1} + C V_{2} = 2 C V$ ⇒ $V_{1} + V_{2} = 2 V$

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A parallel plate capacitor is made by stacking n equally spaced plates connected alternately. If the capacitance between any two plates is C then the resultant capacitance is

(n – 1)C

(n + 1)C

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Explanation

The given arrangement becomes an arrangement of (n – 1) capacitors connected in parallel. So C_R = (n – 1)C

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n identical condensers are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be

Energy and potential difference remain the same

Energy remains the same and the potential difference is nV

Energy increases n times and potential difference is nV

Energy increases n times and potential difference remains the same

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Explanation

According to energy conservation, energy remains the same

⇒ $U_{p a r a l l e l} = U_{s e r i e s}$

⇒ $\frac{1}{2} (n C) V^{2} = \frac{1}{2} (\frac{C}{n}) V'^{2}$

⇒ V' = nV

(V’ = potential difference across series combination)

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Three capacitors of capacitances 3 μF, 9 μF and 18 μF are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases $(\frac{C_{s}}{C_{p}})$ will be

1 : 15

15 : 1

1 : 1

1 : 3

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Explanation

$\frac{1}{C_{s}} = \frac{1}{3} + \frac{1}{9} + \frac{1}{18} = \frac{1}{2}$

⇒ $C_{s} = 2 μ F$

$C_{p} = 3 + 9 + 18 = 30 μ F$

⇒ $\frac{C_{s}}{C_{p}} = \frac{2}{30} = \frac{1}{15}$

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Two capacitances of capacity C₁and C₂ are connected in series and potential difference V is applied across it. Then the potential difference across C₁will be

$V \frac{C_{2}}{C_{1}}$

$V \frac{C_{1} + C_{2}}{C_{1}}$

$V \frac{C_{2}}{C_{1} + C_{2}}$

$V \frac{C_{1}}{C_{1} + C_{2}}$

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Explanation

Charge flowing $= \frac{C_{1} C_{2}}{C_{1} + C_{2}} V$ .

So potential difference across capacitor 1,

$V_{1} = \frac{C_{1} C_{2} V}{C_{1} + C_{2}} \times \frac{1}{C_{1}} = \frac{C_{2} V}{C_{1} + C_{2}}$

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The capacities of two conductors are C₁and C₂ and their respective potentials are V₁and V₁. If they are connected by a thin wire, then the loss of energy will be given by

$\frac{C_{1} C_{2} (V_{1} + V_{2})}{2 (C_{1} + C_{2})}$

$\frac{C_{1} C_{2} (V_{1} - V_{2})}{2 (C_{1} + C_{2})}$

$\frac{C_{1} C_{2} {(V_{1} - V_{2})}^{2}}{2 (C_{1} + C_{2})}$

$\frac{(C_{1} + C_{2}) (V_{1} - V_{2})}{C_{1} C_{2}}$

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Explanation

Initial energy $U_{i} = \frac{1}{2} C_{1} V_{1}^{2} + \frac{1}{2} C_{2} V_{2}^{2}$ ,

Final energy $U_{f} = \frac{1}{2} (C_{1} + C_{2}) V^{2}$ (where $V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$ )

Hence energy loss $Δ U = U_{i} - U_{f} = \frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} - V_{2})}^{2}$

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Three condensers each of capacitance 2F are put in series. The resultant capacitance is

$\frac{3}{2} F$

$\frac{2}{3} F$

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Explanation

$\frac{1}{C} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$

⇒ $C = \frac{2}{3} F$

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2 μF capacitance has potential difference across its two terminals 200 volts. It is disconnected with battery and then another uncharged capacitance is connected in parallel to it, then P.D. becomes 20 volts. Then the capacity of another capacitance will be

2 μF

4 μF

18 μF

10 μF

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Explanation

By using, common potential $V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$

⇒ $20 = \frac{2 \times 200 + C_{2} \times 0}{2 + C_{2}}$ ⇒ C₂ = 18 μF

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