A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±

Question

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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?

Physics

1 answer:

inna [3.1K] · Answer 1 · 2026-03-11T12:59:57+03:00

Answer: A) 2 B) 4 C) 1

Explanation:

The electric field associated with a parallel-plate capacitor is defined as:

A) E=Q/(L^2 * ε0); if we double the charge, the resultant electric field becomes twice as strong as its initial value.

B) Referring to the earlier electric field expression, if the plate size is doubled, the final electric field will be a quarter of the original strength.

C) If the separation distance between the plates is increased twofold, the resulting electric field remains unchanged from its initial state.