The capacitance of the capacitors of plate areas $A_{1}$ and $A_{2} (A_{1} < A_{2})$ at a distance $d$ is

$\frac{ϵ_{0} A_{1}}{d}$
$\frac{ϵ_{0} A_{2}}{d}$
$\frac{ϵ_{0} (A_{1} + A_{2})}{d}$
$\frac{ϵ_{0} \sqrt{A_{1} A_{2}}}{d}$

Question

The capacitance of the capacitors of plate areas A-1- and A-2- -A-1- - A-2- a distance d isdfrac -epsilon-0-A-1-d-dfrac -epsilon-0-A-2-d-dfrac -epsilon-0-A-1- - A-2-d-dfrac -epsilon-0-sqrt -A-1-A-2-d-

Toppr Admin · Accepted Answer

The capacitance of the capacitors of plate areas A1 and A2(A1<A2) at a distance d isϵ0A1dϵ0A2dϵ0(A1+A2)dϵ0√A1A2d

The capacitance of the capacitors of plate areas $A_{1}$ and $A_{2} (A_{1} < A_{2})$ at a distance $d$ is

$\frac{ϵ_{0} A_{1}}{d}$
$\frac{ϵ_{0} A_{2}}{d}$
$\frac{ϵ_{0} (A_{1} + A_{2})}{d}$
$\frac{ϵ_{0} \sqrt{A_{1} A_{2}}}{d}$