- #1

nagyn

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## Homework Statement

Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the quantity of charge you place on capacitor 1 is six times the quantity you place on capacitor 2.

How do the potential differences across each of the two capacitors compare to each other? (V1/V2 = ?)

## Homework Equations

I tried two different approaches:

V(A->B) = -W(A->B)/qtest = -(integral of E*dl from A to B)

V = q/C

C = εA/d

## The Attempt at a Solution

q1 = 6q2

r1 = 2q2

V1/V2 = (kq1/r1^2)*r1 / (kq2/r2^2)*r^2 = (6q2/2r2)/(q2/r2) = 3

This is not correct.

C1/C2 = d1/d2 = 2

V = q/C = 6/2 = 3

I must be misunderstanding potential difference but I'm not sure where I'm going wrong.