Calculation using dielectrics

Question

Given a 7.4 pF air-filled capacitor, you are asked to convert it to a capacitor that can store up to 7.4 mJ with a maximum potential difference of 652 V. Which dielectric in Table 25-1 should you use to fill the gap in the capacitor if you do not allow for a margin of error?

I did the following:

C=(k*e*A)/d and Q=C*V=>C=Q/V

So:

Q/V=(k*e*A)/d=>k=Q/V*d/(A*e)=>k=7.4*10^-6/652*7.4*10^-12=>k=8.7616*10^-20

The answer is 4.7, and uses the potential between capacitors formula. Why are the formulas I used wrong?

Answer 1

Using the formula U = 1/2 C x V², you can get the capacitance necessary to to store 7.4 E -6 J (assuming mj stands for micro-joules).

Solving for C = 2 x 7.4E-6 / 652^2 = 3.48 E-11 F. This is the capacitance needed to store the required amount of energy.

C = k x C_o where C_o is the capacitance without the dielectric. Solve for k = C / C_o.

k = 3.48 E-11 F / 7.4 E -12 F = 4.7