Two identical metal plates are given poistive charges `Q_1` and `Q_2` `(ltQ_1)` respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potencial difference between them is

Options

A. Less than that before the balls touched
B. Greater than that before the balls touched
C. Same as that before the balls touched
D. Zero

Answer

B

Explanation

`F_1=(Kq_1q_2)/(r^2),F_2=(K((q_1+q_2)/2)^(2))/r^2 `
As `((q_1+q_2)/2)^(2) gtq_1q_2 :. F_2 gt F_1`