Initial electric field strength

[ralph344]

Homework Statement

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?

Homework Equations

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

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...

 

[robb_]

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

 

[ralph344]

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

 

[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.