Question

A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of two parallel circular plates with radius 2R and separation of 2d. If the two capacitors have the same charge on them, what is the ratio of the electric fields between the plates of the capacitors, E1/E2?

Answer 1

Answer:

$\frac{E_1}{E_2}= 4$

Explanation:

Capacitance C is given by

$C= \frac{\epsilon_0A}{d}$

A= area of capacitor cross section

d= distance

therefore,

$C_1= \frac{\epsilon_0A_1}{d_1}$

A_1= πR^2

d_1= d

$C_2= \frac{\epsilon_0A_2}{d_2}$

A_= π(2R)^2

d_2 = 2d

$q= \frac{\epsilon_0A_1}{d_1}V_1$

threfore

$V_1= \frac{qd_1}{\epsilon_0A_1}$

and

$V_2= \frac{qd_2}{\epsilon_0A_2}$

also we know that E= V/d

⇒$\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}$

⇒$\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}$

= A_1/A_2= $\frac{4R^2}{R^2}$=4

therefore,

$\frac{E_1}{E_2}= 4$