The electric field between the square plates of a parallel-plate capacitor has magnitude E. The potential across the plates is maintained at constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. (i)The magnitude of the electric field between the plates after being pulled is equal to A) 4E. B) 2E. C) E. D) E/2. E) E/4. (ii) If the plates were charged but not connected to the battery as they were pulled apart to double the separation, which of the above answers give the correct electric field ?

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mavila18 mavila18 · Answer 1 · 2020-05-07T02:19:41+02:00

Answer:

D) E/2

Explanation:

i) To find the magnitude of electric field between the plates you use:

[tex]E=\frac{\Delta V}{d}[/tex]

where V is the voltage and d the separation between plates. If d is doubled you obtain:

[tex]E'=\frac{\Delta V}{2d}=\frac{1}{2}\frac{\Delta V}{d}=\frac{1}{2}E[/tex]

That is, the magnitude of the electric field is halved.

Then, the answer is D.

ii) The magnitude of E' is E/2 because the charge on the plates generates the field E.

Then, the answer is D again