Question

A parallel-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?

Answer 1

Answer: 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.