A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system

  1. Increases by a factor of 2
  2. Increases by a factor of 4
  3. Decreases by a factor of 2
  4. Remains the same

Answer (Detailed Solution Below)

Option 3 : Decreases by a factor of 2
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Detailed Solution

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Concept:

Capacitor is defined as a component in an electrical circuit which has the ability to store energy in the form of an electrical charge producing a potential difference across its plates. It is like rechargeable battery.

Charge on capacitor is given by:

q = CV

where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.

Energy stored in capacitor is calculated using the formula, U:

\(\rm U= \frac{1}{2}C V^2\)

where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.

Calculation:

F1 Savita Others 22-8-22 D15

Charge on capacitor

q = CV

When it is connected with another uncharged capacitor

F1 Savita Others 22-8-22 D17

 \(V_c =\frac{q_1+q_2}{C_1 + C_2}= \frac{q + 0}{C + C}\)

\(\rm V_c= \frac{V}{2}\)

Initial energy

\(\rm U_f = \frac{1}{2}C \left(\frac{V}{2}\right)^2 + \)\(\rm\frac{1}{2}C\left(\frac{V}{2}\right)^2\)

\(= \rm\frac{CV^2}{4}\)

Loss of energy  = Ui - Uf

\(\rm=\frac{CV^2}{4}\)

i.e. decreases by a factor (2)

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