- #1

knowlewj01

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## 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 [itex]\pm q[/itex].

(a) What is the electric field strength [itex]E(q,A)[/itex] inside the capacitor?

(b) By how many percent does the electric field strength change if the positive charge is doubled?

## Homework Equations

[itex]\oint\bf{E.n} dS = 4\pi k Q[/itex]

Electric Field inside a capacator: [itex]E = \frac{Q}{A \epsilon_0}[/itex]

## 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 [itex]\oint\bf{E.n} dS[/itex]

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)

[itex]\oint\bf{E.n} dS = 4\pi k Q = \frac{q}{\epsilon_0}[/itex]

[itex]E A = \frac{q}{\epsilon_0}[/itex]

[itex]E = \frac{q}{A \epsilon_0}[/itex]

this is in agreement with the given formula i found.

(b) double the positive charge:

[itex]\oint\bf{E.n} dS = 4\pi k Q = \frac{2q}{\epsilon_0}[/itex]

[itex]E = \frac{2q}{A \epsilon_0}[/itex]

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.