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

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Homework Help Overview

The discussion revolves around the effect of doubling the charge on the electric field strength in a parallel-plate capacitor. The original poster presents a problem involving the relationship between charge and electric field strength, specifically asking for the ratio of final to initial electric field strengths when the charge is doubled.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the formula for electric field strength but seems uncertain about its applicability to a parallel-plate capacitor. Some participants question the use of the point charge formula in this context and suggest considering the constant electric field inside the capacitor.

Discussion Status

The discussion has seen some exploration of the relationship between charge and electric field strength, with participants providing guidance on the correct formula to use for a parallel-plate capacitor. There is a mix of interpretations regarding the effect of doubling the charge, and while one participant claims to have found the correct ratio, the discussion remains open to further clarification.

Contextual Notes

Participants are navigating the differences between formulas applicable to point charges versus those for parallel-plate capacitors, indicating potential confusion about the assumptions underlying the problem setup.

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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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I tried 1/2 but it was incorrect. Help please!
 
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 the voltage and d is the distance between the plates.
 
i figured it out. it wasnt 1/2, for Efinal/Einitial = 2
 

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