Initial electric field strength

AI Thread Summary
The initial electric field strength between oppositely charged parallel plates is 3.0 X 10^3 N/C. When half the charge is removed and the plate separation decreases from 12 mm to 8 mm, the expected electric field strength calculation leads to confusion. The correct approach involves using the surface charge density formula, E = n/ε₀, rather than the point charge equation. The final calculated electric field strength is 1.5 X 10^3 N/C, highlighting the importance of using the appropriate equations for parallel plate capacitors. Understanding the distinction between point charge and parallel plate equations is crucial for accurate calculations.
ralph344
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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...
 
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The electric field equation you state is for a point charge. ( and other limited cases)
 
But that's the only equation taught in this section... there is no other equations... anymore hints?
 
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.
 
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