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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. 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?

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jorgezarate

Answer:

E(final)/E(initial)=2

Explanation:

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

[tex]E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\[/tex]

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

E(final)/E(initial)=2

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