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
Updated On: Jul 7, 2022
$6 \times 10^{7} \, V$
$3 \times 10^{7} \, V$
$4 \times 10^{7} \, V$
$2 \times 10^{7} \, V$
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The Correct Option isA
Solution and Explanation
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$
The total capacitance of this equivalent single capacitor depends both on the individual capacitors and how they are connected. There are two simple and common types of connections, called series and parallel, for which we can easily calculate the total capacitance.
When one terminal of a capacitor is connected to the terminal of another capacitors , called series combination of capacitors.
Capacitors in Parallel
Capacitors can be connected in two types which are in series and in parallel. If capacitors are connected one after the other in the form of a chain then it is in series. In series, the capacitance is less.
When the capacitors are connected between two common points they are called to be connected in parallel.
When the plates are connected in parallel the size of the plates gets doubled, because of that the capacitance is doubled. So in a parallel combination of capacitors, we get more capacitance.
