A parallel-plate capacitor has capacitance C = 12.5 pF when the volume between t

Question

A parallel-plate capacitor has capacitance C = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted the charge on each plate has magnitude 45.0 pC.

1.) What is the electric field at a point midway between the plates before the dielectric has been inserted?

2.) What is the electric field at a point midway between the plates after the dielectric has been inserted?

I know that the dielectric constant is 1.8 and the potential difference between the plates with and without the dielectric is 2.0 V. I just don't understand how to find the electric field? And would they be the same?

Explanation / Answer

Electric field is given by:

E = V/d

Since potential difference between the plates is equal to 2.0 V, with and without the dielectric, So electric field will also be same in both case

[we know that E = kq/d^2 . and V = kq/d, from these relations]

Now Part 1 & 2.

E = V/d

C = e0*A/d

d = e0*A/C = 8.85*10^-12*pi*0.03^2/(12.5*10^-12)

d = 0.002

So,

E = 2/0.002 = 1000 V/m (In both cases with and without dielectric)

Please Upvote.

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