Question

15. If electric field between plates of parallel plate
capacitor is 2 N/C and charge on two plates are
10 C and 3 C then force on one of the plates is
(1) 20 N
(2) 13 N
(3) 60/7N
(4)7/2N

Answer 1

Answer: (3)60/7 N

Explanation:

Please find the explanation in the attachment

Answer 2

Answer:

60/7 N.

Explanation:

Since the electric field between the parallel plates is 2 so, taking the electric fields to be E1 and E2 hence E1 - E2 will be 2.

Now, E = q1/2Aε - q2/2Aε where A is the area of the capacitor.

So, putting the values we will get E = 10/2Aε - 3/2Aε so on solving we will get the value of the q2 to be 2Aε/3.5.

Now, we know that the force on 2 due to 1 will be electric field on 1 per charge on 2 hence, F = E1 * q2 = q1/2Aε *q2 which on solving we will get the value of the F to be 60/7.