Question

Five identical conducting plates $1,2,3,4$ and $5$ are fixed parallel to and equidistant from each other (see figure). Plates $2$ & $5$ are connected by a conductor while $1$ & $3$ are joined by another conductor. The junction of $1$ & $3$ and the plate $4$ are connected to a source of constant e.m.f. $V_0$. Find:
(i) The effective capacity of the system between the terminals of the source.
(ii) the charges on plates $3$ & $5$
Given $d=$ distance between any successive plates & $A=$ area of either face of each plate

Answer 1

Let $q$ be the charges given to the system,

$q_3+q_4=q⟶\left(i\right)$

$q_1+q_2+q_3=q⟶\left(ii\right)$

$q_1-q_2=0\Rightarrow q_1-q_2⟶\left(iii\right)$

$q_4-q_3+q_2=0$

$q_4+q_2=q_3⟶\left(iv\right)$

on solving the above equations

$q_3=\frac{3q}{5}⟶\left(v\right)$

so, $△V=\frac{3q}{5c}$ $[c=$Capacitance of each capacitor$]$

$\left(i\right)$ so, capacitance $=\frac{Q}{△V}=\frac{q}{\frac{3q}{5c}}=\frac{5c}{3}=\frac{5{\epsilon }_{o}A}{3d}$

$\left(ii\right)$ Charge on Plate $3\Rightarrow q_2$

By solving the equations $\left(i\right),\left(iii\right),\left(v\right)$

$q_2=\frac{-q}{10}$

Now we have,

$△V=\frac{3q}{5c}$,so,$V_o=\frac{3q}{5c}$

$\Rightarrow \frac{5V_{o}c}{3}=q$

$\Rightarrow \frac{5v_o}{3}.\frac{A{\epsilon }_o{V}_o}{6d}$

And charge on plate $5=q_4$

$q_3+q_4=q$

$q_4=q-\frac{3q}{5}=\frac{2q}{5}$

$=\frac{2}{5}\times \frac{5v_{o}A{\epsilon }_o}{3d}$

$=\frac{2}{3}\frac{v_{o}A{\epsilon }_o}{d}$