Aparallel-plate capacitor is constructed of two square plates, size l×l, separated by distance d. the plates are given charge ±q.
part a
what is the ratio ef/ei of the final electric field strength ef to the initial electric field strength ei if qis doubled?
part bwhat is the ratio ef/ei of the final electric field strength ef to the initial electric field strength ei if lis doubled? part cwhat is the ratio ef/ei of the final electric field strength ef to the initial electric field strength ei if d is doubled?

A) 2 B) 4 C) 1

Explanation:

The Electric field from a parallel-plate capacitor  is given by:

A) E=Q/(L^2 * ε0) so if we put a charge double the final electric field is double that the original.

B) from the above expression for the electric field,  If the size of the plate is double, then the E final is four times weaker that the original.

C) If the distante between plates is doubled the final electric field is the same that initial.

are there answer choices to this question? this seems like a question where you have options to choose from.

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