Question:

A capacitor is charged with a battery and energy stored is \(U\). After disconnecting battery another capacitor of same capacity is connected in parallel with it. Then energy stored in each capacitor is:

Updated On: Apr 21, 2025

\(\frac U2\)
\(\frac U4\)
\(4U\)
\(2U\)

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The Correct Option is B

Solution and Explanation

Parallel combination of Capacitor

\(Energy\ stored\ (U) =\frac {q^2}{2C}\)

After connecting with another capacitor

\(V_{common}=\frac {q_1+q_2}{C_1+C_2}=\frac {q+0}{C+C}=\frac {q}{2C}\)

∴Energy on each capacitor =\(\frac {1}{2} CV^2_{common}\) = \(\frac 12 (\frac {q}{2C})^2\) = \(\frac U4\)

So, the correct option is (B): \(\frac U4\)

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Concepts Used:

Electrostatic Potential and Capacitance

Electrostatic Potential

The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.

Some major things that we should know about electric potential:

They are denoted by V and are a scalar quantity.
It is measured in volts.

Capacitance

The ability of a capacitor of holding the energy in form of an electric charge is defined as capacitance. Similarly, we can also say that capacitance is the storing ability of capacitors, and the unit in which they are measured is “farads”.

The capacitor is in Series and in Parallel as defined below;

In Series

Both the Capacitors C₁ and C₂ can easily get connected in series. When the capacitors are connected in series then the total capacitance that is C_totalis less than any one of the capacitor’s capacitance.

In Parallel

Both Capacitor C₁ and C₂ are connected in parallel. When the capacitors are connected parallelly then the total capacitance that is C_total is any one of the capacitor’s capacitance.