Question

A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±Q. Let's consider how the electric field changes if one of these variables is changed while the others are held constant. Part A What is the ratio E(final)/E(initial) of the final to initial electric field strengths if Q is doubled?

Answer 1

E(final)/E(initial)=2

Step-by-step explanation:

Applying the law of gauss to two parallel plates with charge density equal σ:

`E=\sigma/\epsilon_(o)=Q/(L^(2)*\epsilon_(o))\\`

So, if the charge is doubled the Electric field is doubled too

E(final)/E(initial)=2