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

Answer:

Explanation:

Given

Area of capacitor Plates $A=L\times L$

distance between plates is d

capacitance C is given by

$C=\frac{\epsilon A}{d}$

$C=\frac{\epsilon \cdot L^2}{d}$

Provided V is Voltage

$Charge(Q)=capacitance(C)\times Voltage(V)$

If L is doubled

Capacitance $C'=\frac{\epsilon \cdot (2L)^2}{d}$

$C'=4\times \frac{\epsilon \cdot L^2}{d}$

Electric field is given by

$E=\frac{Q}{\epsilon _0A}$

$E_i=\frac{Q}{\epsilon _0L^2}---1$

$E_f=\frac{Q}{\epsilon _0(2L)^2}---2$

divide 1 and 2 we get

$\frac{E_i}{E_f}=\frac{(2L)^2}{L^2}$

$\frac{E_f}{E_i}=\frac}{1}{4}$