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MCQEasyJEE 2023Capacitors & Dielectrics

JEE Physics 2023 Question with Solution

A parallel plate capacitor of capacitance 2 F2 \, \text{F} is charged to a potential VV. The energy stored in the capacitor is E1E_1. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is E5E_5. The ratio E5/E1E_5 / E_1 is:

  • A

    2:32 : 3

  • B

    1:21 : 2

  • C

    1:41 : 4

  • D

    2:12 : 1

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A capacitor of capacitance C=2 FC = 2 \, \text{F} is charged to potential VV. Its initial energy is E1E_1. It is connected in parallel to another identical uncharged capacitor.

Find: The ratio E5/E1E_5/E_1.

The energy stored in a capacitor is

E=12CV2E = \frac{1}{2} C V^2

For the initial single capacitor,

E1=12×2×V2=V2E_1 = \frac{1}{2} \times 2 \times V^2 = V^2

When two identical capacitors are connected in parallel, the total capacitance becomes

Ctotal=C+C=2C=4 FC_{\text{total}} = C + C = 2C = 4 \, \text{F}

Using the same potential VV as shown in the provided solution,

E5=12×4×V2=2V2E_5 = \frac{1}{2} \times 4 \times V^2 = 2V^2

Therefore,

E5E1=2V2V2=2\frac{E_5}{E_1} = \frac{2V^2}{V^2} = 2

So the ratio is 2:12 : 1. The correct option is D.

Common mistakes

  • Assuming the capacitance remains unchanged after connecting the second capacitor is incorrect, because in parallel combination capacitances add. Use Ctotal=C1+C2C_{\text{total}} = C_1 + C_2.

  • Using the wrong energy formula is a common error. The correct expression is E=12CV2E = \frac{1}{2}CV^2, not CV2CV^2.

  • Confusing series and parallel combinations leads to an incorrect equivalent capacitance. Here the capacitors are in parallel, so the equivalent capacitance increases, not decreases.

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