# Parallel plate capacitor with dielectric. Find capacitance

## Homework Statement

The plates of a parallel plate capacitor measure 15cm x 15cm, and they are separated by 6cm. The capacitor is charged with a potential difference of 18V, then removed from the voltage source and kept isolated. A 15cm x 15cm x 3cm slab of dielectric material (k = 2.5) is inserted between the plates. What is the capacitance in µF after the dielectric is inserted?

Vi=Ei*d (1)
E=(ke)Q/r^2 (2)
V=(Vi)/k (3)
C=Q/V (4)

## The Attempt at a Solution

I used the formulas above in that order. I solved first for E(initial) using V(initial) and d=.06. I got E(i)=300
I then plugged E(i) it into eqn(2) to get Q. Q=1.2 x 10^-10
Next, I found V using eqn(3): 18V/2.5 to get V=7.2

Finally, I plugged Q and V into eqn 4 and got 1.7 x 10^-11

This doesn't match any of my multiple choice answers. Can someone help me figure out where I went wrong or give me a better plan of attack for this type of problem?

rl.bhat
Homework Helper
What is the capacitance in µF after the dielectric is inserted?

Capacitance does not depend on the charge on the plate or PD across the plates. When you introduce dielectric slab, the capacitance increases.

C = k*Co where C is the capacitance with dielectric, Co is the capacitance without dielectric and k is th dielectric constant.

Does the size of the dielectric affect the answer?
C for parallel plates= (eps)A/d
I used that to get C(i) and then multiply by the dielectric coefficient, but I am still incorrect

rl.bhat
Homework Helper
Does the size of the dielectric affect the answer?
C for parallel plates= (eps)A/d
I used that to get C(i) and then multiply by the dielectric coefficient, but I am still incorrect
If the dielectric is completely occupies the space between the plates, then the C = Co*k applies. If the a part of the space is filled, them

C = εο*A /[(d-x) + x/k], where x is the thickness of the dielectric slab.