Question

Each of the plates shown in the figure has surface area $\left(96/{ϵ}_0\right)\times {10}^{-12}$ Fm on one side and the separation between the consecutive plates is $4.0mm$. The emf of the battery connected is $10$ volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.

Answer 1

Here three capacitors are formed

And each of

Surface area of the plates is $A=\frac{96}{{\epsilon }_0}\times {10}^{-12}f.m$

The distance between the plates is $d=4mm=4\times {10}^{-3}m$

$\therefore$ Capacitance of a capacitor

$C=\frac{{\epsilon }_{0}A}{d}=\frac{{\epsilon }_0\frac{96\times {10}^{-12}}{{\epsilon }_0}}{4\times {10}^{-3}}=24\times {10}^{-9}F$

$\therefore$ As three capacitor are arranged is series

So, $C_{eq}=\frac{C}{q}=\frac{24\times {10}^{-9}}{3}=8\times {10}^{-9}$

$\therefore$ The total charge to a capacitor

$q=CV$

$=8\times {10}^{-9}\times 10=8\times {10}^{-8}C$

$\therefore$ The charge of a single Plate $q^{\prime }=2\times 8\times {10}^{-8}=16\times {10}^{-8}=0.16\times {10}^{-6}=0.16\mu C$