Question

A, b, c, d, e and f are conducting plates each f area a, and any two consecutive plates separated by a distance d. The net energy stored in the system after the switch is 's' closed is

Answer 1

Answer:

follow me

Explanation:

plzz i will follow u

Answer 2

(The figure missing in the question is attached below)

Given:

Area of the conducting plates = A

Distance between the plates = d

To find:

The net energy stored in the system after the swith ‘s’ is closed

Calculation:

The plates A and C are connected, so the potential difference between the plates will be zero

           $V_{A}=V_{C}$     ……(1)

The plates C and E are connected, so the potential difference between the plates will be zero

           $V_{C}=V_{E}$     ……(2)

 From equations (1) & (2)

           $V_{A}=V_{B}$

Therefore the net capacitance between the plates A and E will be zero.

The effective capacitance between the plates E and F is

           $C=\frac{A\epsilon_{0}}{d}$

The energy stored in the capacitor is given by the formula

           $U=\frac{1}{2} CV^{2}$

           $U=\frac{A\epsilon_{0}}{2d} V^{2}$

The net energy stored in the system after the swith ‘s’ is closed is  $U=\frac{A\epsilon_{0}}{2d} V^{2}$