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?

Question

Madeleine Carney

Physics

5 January, 12:33

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?

+5

Answers (1)

Know the Answer?

Lorena Parks · 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.