Question

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

a. What is the ratio Ef/Ei of the final to initial electric field strengths if L is doubled?

Answer 1

Step-by-step explanation:

Given

Area of capacitor Plates
`A=L* L`

distance between plates is d

capacitance C is given by

`C=(\epsilon A)/(d)`

`C=(\epsilon \cdot L^2)/(d)`

Provided V is Voltage

`Charge(Q)=capacitance(C)* Voltage(V)`

If L is doubled

Capacitance
`C'=(\epsilon \cdot (2L)^2)/(d)`

`C'=4* (\epsilon \cdot L^2)/(d)`

Electric field is given by

`E=(Q)/(\epsilon _0A)`

`E_i=(Q)/(\epsilon _0L^2)---1`

`E_f=(Q)/(\epsilon _0(2L)^2)---2`

divide 1 and 2 we get

`(E_i)/(E_f)=((2L)^2)/(L^2)`

`(E_f)/(E_i)=\frac}{1}{4}`