# Electric field strength

1. Apr 8, 2007

### ralph344

1. The problem statement, all variables and given/known data
The initial electric field strength between oppositely charged parallel plates is 3.0 X 10^3 N/C, what would the electric field strength become if half of the charge were removed from each plate and the separation of the plates were changed from 12 mm to 8 mm?

2. Relevant equations
E= kq/r^2
k=9.0 X 10^9

3. The attempt at a solution
I used proportions.
Since the formula is E=kq/r^2, i rearranged the formula to Er^2/q = k. Since the k value is going to stay constant for both.
Er^2/q = Er^2/q
[(3.0 X 10^3)(0.012)^2]/q=[E(0.008)^2]/[0.5q]
E= 3.5 X 10^3 N/C
But the answer is 1.5 X 10^3 N/C...

2. Apr 8, 2007

### robb_

The electric field equation you state is for a point charge. ( and other limited cases)

3. Apr 8, 2007

### ralph344

But that's the only equation taught in this section... there is no other equations... anymore hints?

4. Apr 9, 2007

### Jide

hey! i solved your problem! the equation you are using is that of a point charge like robb said. the electric field strength in a parallel plate capacitor is given by E=n/epsilon zero. n is the surface charge density of the plates which are equal in magnitude. the surface charge density is simply the amount charge on the plates divided by its surface area.