Spherical Capacitors

Practice Questions:

Question 1: easy

Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

1. 8 times

2. 4 times

3. 2 times

4. 32 times

View Answer

Let \(r\) be the radius of a small drop and \(R\) be the radius of the big drop. Volume conservation: \(8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 ⇒ R = 2r\). Capacitance of a spherical drop is \(C = 4\pi\epsilon_0 R\). So, \(C_{\text{big}} = 4\pi\epsilon_0 (2r) = 2 (4\pi\epsilon_0 r) = 2 C_{\text{small}}\).

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Question 2: easy

Two metallic charged spheres whose radii are \( 20\text{cm} \) and \( 10\text{cm} \) respectively, have each \( 150\ \mu\text{C} \) positive charge. The common potential after they are connected by a conducting wire is

1. \( 9 \times 10^6\text{ volts} \)

2. \( 4.5 \times 10^6\text{ volts} \)

3. \( 1.8 \times 10^7\text{ volts} \)

4. \( 13.5 \times 10^6\text{ volts} \)

View Answer

The total charge is \( Q = 150 \mu\text{C} + 150 \mu\text{C} = 300 \mu\text{C} \). The total capacitance is \( C = 4piepsilon_0(R_1 + R_2) = \frac{0.3}{9 \times 10^9}\text{ F} \). The common potential is \( V = \frac{Q}{C} = \frac{300 \times 10^{-6} \times 9 \times 10^9}{0.3} = 9 \times 10^6\text{ V} \).

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