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%