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## 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?

## Homework Equations

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?