How Does Doubling Charge Affect Electric Field Strength in a Capacitor?

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In summary, A parallel-plate capacitor with two square plates of size L x L and separated by distance d has a constant electric field inside, given by E=V/d. If the charge +Q on the plates is doubled, the ratio of final to initial electric field strengths is 2.
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phyBnewb
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Homework Statement


A parallel-plate capacitor consists of two square plates, size L x L, separated by distance d. The plates are given charge +-Q . Each part changes only one quantity; the other quantities have their initial values.

Part A

What is the ratio E(final)/E(initial) of the final to initial electric field strengths if Q is doubled?





Homework Equations



I know that:E = kQ/r^2 ,where k= 8.99*10^9


The Attempt at a Solution



I am not sure where to start here for this question.
Since it is a ratio of E final / E initial would Q cause E to be something like this?
E = k(2Q)/r^2 => 1/2E for E final?
 
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  • #2
I tried 1/2 but it was incorrect. Help please!
 
  • #3
You are using the formula for a point charge in the case of a parallel plate capacitor.
The electric field inside the capacitor is constant and is given by E=V/d where V is teh voltage and d is the distance between the plates.
 
  • #4
i figured it out. it wasnt 1/2, for Efinal/Einitial = 2
 

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