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Q. Two parallel conducting plates of area $A = 2 .5 \,m^{2}$ each are placed $6\, mm$ apart and are both earthed. A third plate, identical with the first two, is placed at a distance of $2\, mm$ from one of the earthed plates and is given a charge of $1\, C$. The potential of the central plate is
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Electrostatic Potential and Capacitance

Solution:

The plates $1$ and $2$ will form one capacitor and plates $2$ and $3$ will form another capacitor. These two capacitors are in parallel. Their effective capacitance is
$C=\frac{\varepsilon_{0}A}{d}+\frac{\varepsilon_{0}A}{2d}=\frac{3}{2} \frac{\varepsilon_{0}A}{d}$
Potential of central plate, $V=\frac{Q}{C}=\frac{Q2d}{3\varepsilon_{0}A}$
$=\frac{1\times2\times2\times10^{-3}}{3\times8.85\times10^{-12} \times2.5 } \approx6 \times10^{7}\, V$