Question

Five identical conducting plates, $1.2,3,4$ and $5$ are fixed parallel plates equidistant from each other (see figure). A conductor connects plates $2$ and $5$ while another conductor joins $1$ and $3$. The junction of $1$ and $3$ and the plate $4$ are connected to a source of constant emf $V_o$. Find the charges on the plates $3$ and $5$. Given, $d=$ distance between any two successive plates and $A=$ area of either face of each plate.

• A. $q_3=\frac{2}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right),q_5=\frac{4}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right)$
• B. $q_3=\frac{4}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right),q_5=\frac{2}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right)$
• C. $q_3=\frac{4}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right),q_5=\frac{4}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right)$
• D. $q_3=\frac{2}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right),q_5=\frac{2}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right)$

Answer 1

The correct option is B $q_3=\frac{4}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right),q_5=\frac{2}{3}\left(\frac{{ϵ}_{o}A{V}_o}{d}\right)$

The effective capacity has to be found between $V_1$ and $V_2$.

$q_3=+\left[C\right(V_1-V_1)+C(V_1-V_3\left)\right]$

$=\frac{{ϵ}_{o}A}{d}[V_o+\frac{V_o}{3}]=\frac{4{ϵ}_{o}A{V}_o}{3d}$

$q_5=C(V_3-V_2)=\left(\frac{{ϵ}_{o}A}{d}\right)\left(\frac{2V_o}{3}\right)$

$=\frac{2{ϵ}_{o}A{V}_o}{3d}$