Application of Gauss' Laws: Large area parallel plate capacitor

[knowlewj01]

Homework Statement

The plates of a large area parallel plate capacitor of area A are separated by a short
distance. The plates carry an equal but opposite charge $\pm q$.
(a) What is the electric field strength $E(q,A)$ inside the capacitor?
(b) By how many percent does the electric field strength change if the positive charge is doubled?

Homework Equations

$\oint\bf{E.n} dS = 4\pi k Q$

Electric Field inside a capacator: $E = \frac{Q}{A \epsilon_0}$

The Attempt at a Solution

the electric field inside a parallel plate capacator is uniform so it does not matter where we do the surface integral $\oint\bf{E.n} dS$
i choose to do the integral on a plane equal in area to the area of the capacators, i will call this area A.

(a)
$\oint\bf{E.n} dS = 4\pi k Q = \frac{q}{\epsilon_0}$

$E A = \frac{q}{\epsilon_0}$

$E = \frac{q}{A \epsilon_0}$

this is in agreement with the given formula i found.

(b) double the positive charge:

$\oint\bf{E.n} dS = 4\pi k Q = \frac{2q}{\epsilon_0}$

$E = \frac{2q}{A \epsilon_0}$

so this is twice that of in the first case. is this the right way to look at this? i thought what would be the difference if i changed the negative one instead, since in this solution i am ignoring the fact that there is a negative charge at all. i think that it would have the same effect.

 

[darkys]

the percent change can be worked out as:percent change = (\frac{2q}{A \epsilon_0} - \frac{q}{A \epsilon_0})/ \frac{q}{A \epsilon_0} * 100percent change = 100%