Question

Two identical thin metal plates has charge q1 and q2 respectively such that q1 > q2. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is

• A. (q1+q2/C)
• B. (q1-q2/C)
• C. (q1-q2/2C)
• D. 2(q1-q2/C)

Answer 1

Answer: (C)

(q1-q2/2C)

Answer 2

E=q1−q2/2ε0A$E=\frac{q_1−q_2}{2ε_0A}$
$v=\frac{(q1−q2)d}{2ε0A}$
= $\frac{q_1-q_2}{2C}$
So, the correct option is (C): (q1-q2)/2C