Answer :
Answer:
Explanation:
Given
Area of capacitor Plates [tex]A=L\times L[/tex]
distance between plates is d
capacitance C is given by
[tex]C=\frac{\epsilon A}{d}[/tex]
[tex]C=\frac{\epsilon \cdot L^2}{d}[/tex]
Provided V is Voltage
[tex]Charge(Q)=capacitance(C)\times Voltage(V)[/tex]
If L is doubled
Capacitance [tex]C'=\frac{\epsilon \cdot (2L)^2}{d}[/tex]
[tex]C'=4\times \frac{\epsilon \cdot L^2}{d}[/tex]
Electric field is given by
[tex]E=\frac{Q}{\epsilon _0A}[/tex]
[tex]E_i=\frac{Q}{\epsilon _0L^2}---1[/tex]
[tex]E_f=\frac{Q}{\epsilon _0(2L)^2}---2[/tex]
divide 1 and 2 we get
[tex]\frac{E_i}{E_f}=\frac{(2L)^2}{L^2}[/tex]
[tex]\frac{E_f}{E_i}=\frac}{1}{4}[/tex]