Two identical thin metal plates has charge $\mathrm{q}_1$ and $\mathrm{q}_2$ respectively such that $q_1>q_2$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :
Select the correct option:
A
$\frac{\left(q_1+q_2\right)}{c}$
B
$\frac{\left(q_1-q_2\right)}{C}$
C
$\frac{\left(q_1-q_2\right)}{2 C}$
D
$\frac{2\left(q_1-q_2\right)}{C}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
electric field between plates
$$
\begin{aligned}
& E=\frac{q_1-q_2}{2 A \epsilon_0} \\
& V=E d=\frac{q_1-q_2}{2 A \epsilon_0} d \\
& V=\frac{q_1-q_2}{2 C}
\end{aligned}
$$
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
