Guass' Law problem?

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The electric field near the surface of a uniformly charged conducting plane can be determined by Gauss's law (look it up). If the plane is a square of sides 12.5 cm and carries a charge of 26.5 nC (that's nanocoulombs), what is the magnitude of the electric field near the surface?

Now let's say that a plane with a charge of equal magnitude but opposite sign is brought near the first. What is the force between the two planes?

Now we also know that the charge on a pair of parallel plates is proportional to the potential difference V, namely: q=CV. We also know that the capacitance of two parallel plates is given by : C=?0A/d, where ?0=8.85×10-12 is the permittivity of free space and d is the separation between the plates. Combine these equations and to get the force as a function of the potential. So now if you were to make a graph of force (F) vs. A2/d2, the slope of the graph would be
A) ?0A/2
B) C
C) E
D) ?0
So now let's say that you find a slope of 6.154×10-14 for square plates of width 12.5 cm. What do you get for the value of the permittivity of free space?

If the uncertanty in the slope is 4.738×10-15, is the measured value of ?0 within experimental uncertainty of the accepted value? (Enter Y for yes and N for no)
 
Gauss' law states that the electric field is
E=Q/?0A
just plug in the values in SI units.

The force between the two planes of charge
Voltage =work done /charge = F d/ Q = E d
Plug in the numbers

C = ?0A/d = Q/V
 
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