Question:

See the diagram. Area of each plate is $2.0\,m^2$ and $d = 2\times 10^{-3} C$ is given to $Q$.Then the potential of $Q$ becomes

Updated On: Feb 23, 2024

13 V
10 V
6.67 V
8.825 V

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The Correct Option is C

Solution and Explanation

In the given arrangement, plate $Q$ is common for two capacitors which are connected in parallel. $\therefore \quad C_{ eff }=C_{P}=C_{1}+C_{2}$ $ \Rightarrow C_{P}=\frac{\varepsilon_{0} A}{d}+\frac{\varepsilon_{0} A}{2 d}=\frac{2 \varepsilon_{0} A}{2 d}$ The given arrangement can be shown as Let $V$ be the potential difference across the capacitor, which is equal to the potential of the plate $Q$

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Concepts Used:

Combination of Capacitors

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.

Read Also: Combination of Capacitors

Series capacitors

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.

Read More: Types of Capacitors