Question

Two capacitors of capacitance 2$\mu$F and 3$\mu$F are joined in series. Outer plate first capacitor is at 1000 volt and outer plate of second capacitor is earthed (grounded). Now the potential on inner plate of each capacitor will be

• A. 700 Volt
• B. 200 Volt
• C. 600 Volt
• D. 400 Volt

Answer 1

Answer: (D)

400 Volt

Answer 2

Let the two capacitors be $C_1$ and $C_2$, with capacitances $C_1 = 2 \mu$F and $C_2 = 3 \mu$F. They are connected in series. Let the plates of the first capacitor be A (outer) and B (inner). Let the plates of the second capacitor be C (inner) and D (outer). The outer plate of the first capacitor (A) is at a potential $V_A = 1000$ V. The outer plate of the second capacitor (D) is earthed, so its potential is $V_D = 0$ V. In a series connection, the inner plate of the first capacitor (B) is connected to the inner plate of the second capacitor (C). Let the potential of this common connection be $V$. So, $V_B = V_C = V$.

The potential difference across the first capacitor $C_1$ is $V_1 = V_A - V_B = 1000 - V$. The charge on the first capacitor is $Q_1 = C_1 V_1 = C_1 (1000 - V)$.

The potential difference across the second capacitor $C_2$ is $V_2 = V_C - V_D = V - 0 = V$. The charge on the second capacitor is $Q_2 = C_2 V_2 = C_2 V$.

In a series combination of capacitors, the charge on each capacitor is the same, i.e., $Q_1 = Q_2$. Therefore, we have: $C_1 (1000 - V) = C_2 V$

Substitute the given values of $C_1$ and $C_2$: $2 \mu\text{F} \times (1000 - V) = 3 \mu\text{F} \times V$ $2 (1000 - V) = 3 V$ $2000 - 2V = 3V$ $2000 = 3V + 2V$ $2000 = 5V$ $V = \frac{2000}{5}$ $V = 400$ V

The potential on the inner plate of each capacitor is $V_B = V_C = V = 400$ V.