Question

See the diagram. Area of each plate is $2.0\,m^2$ and $d = 2\times 10^{-3} C$ is given to $Q$.Then the potential of $Q$ becomes

• A. 13 V
• B. 10 V
• C. 6.67 V
• D. 8.825 V

Answer 1

Answer: (C)

6.67 V

Answer 2

In the given arrangement, plate $Q$ is common for two capacitors which are connected in parallel. $\therefore \quad C_{ eff }=C_{P}=C_{1}+C_{2}$ $\Rightarrow C_{P}=\frac{\varepsilon_{0} A}{d}+\frac{\varepsilon_{0} A}{2 d}=\frac{2 \varepsilon_{0} A}{2 d}$ The given arrangement can be shown as Let $V$ be the potential difference across the capacitor, which is equal to the potential of the plate $Q$