Question

A parallel-plate capacitor is charged and then disconnected from the battery, so that the charge Q on its plates cant change. Originally the separation between the plates of the capacitor is d and the electrical field between the plates of the capacitor is E = 6.0 × 10^4 N/C If the plates are moved closer together, so that their separation is halved and becomes d/2, what then is the electrical field between the plates of the capacitor? What if the battery is left connected?

Answer 1

`E_f = (V)/((d)/(2))= 2 (V)/(d)= 2* 6x10^4 (N)/(C) = 1.2x10^5 (N)/(C)`

Step-by-step explanation:

For this case we know that the electric field is given by:

`E= 6 x 10^4 (N)/(C)`

And we want to find the final electric field assuming that the separation is halved and becomes d/2.

For this case we can use the following two equations:

`C = \epsilon_o (A)/(d) = (Q)/(V)` (1)

`E = (\sigma)/(\epsilon_o)` (2)

Where Q represent the charge, V the voltage, d the distance, A the area.

We can rewrite the equation (2) like this:

`E = (\sigma)/(\epsilon_o) = (Q)/(A \epsilon_o)` (3)

And we can solve for Q from equation (1)like this:

`Q = (\epsilon_o A V)/(d)`

And if we replace into equation (3) the previous result we got:

`E = (\epsilon_o A V)/(A d \epsilon_o) = (V)/(d)`

And since the the electric field not change and the distance would be the half we have that the final electric field would be given by:

`E_f = (V)/((d)/(2))= 2 (V)/(d)= 2* 6x10^4 (N)/(C) = 1.2x10^5 (N)/(C)`