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A parallel-plate capacitor consists of two square plates, size L×L, separated by distance d. The plates are given charge ±Q . What is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if:
a. Q is doubled?
b. L is doubled?
c. d is doubled?

Respuesta :

hopemichael95

Answer:

Using

A. .E = σ/εo = (q/A)/εo = = q/Aεo so if q = 2q, then

Ef/Ei = 2

B. If L is 2L then Ef = q/4Aεo and

Ef/Ei = 1/4

C. The electric field strength is not effected by d and as long as σ is unchanged, Ef/Ei = 1

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