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MCQEasyJEE 2025Capacitors & Dielectrics

JEE Physics 2025 Question with Solution

A parallel-plate capacitor of capacitance 40 μF40\,\mu \text{F} is connected to a 100 V100\,\text{V} power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K=2K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are -

  • A

    2 mC2\,\text{mC} and 0.2 J0.2\,\text{J}

  • B

    8 mC8\,\text{mC} and 2.0 J2.0\,\text{J}

  • C

    4 mC4\,\text{mC} and 0.2 J0.2\,\text{J}

  • D

    2 mC2\,\text{mC} and 0.4 J0.4\,\text{J}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Capacitance of the parallel-plate capacitor is C=40×10−6 FC = 40 \times 10^{-6} \, \text{F}, applied voltage is V=100 VV = 100 \, \text{V}, and dielectric constant is K=2K = 2.

Find: The extra charge supplied by the battery and the change in electrostatic energy after inserting the dielectric while the power supply remains connected.

When the battery remains connected, the potential difference stays constant. The new capacitance becomes

C′=KC=2×40×10−6=80×10−6 FC' = KC = 2 \times 40 \times 10^{-6} = 80 \times 10^{-6} \, \text{F}

Initial charge:

Q1=CV=40×10−6×100=4×10−3 C=4 mCQ_1 = CV = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3} \, \text{C} = 4 \, \text{mC}

Final charge:

Q2=C′V=80×10−6×100=8×10−3 C=8 mCQ_2 = C'V = 80 \times 10^{-6} \times 100 = 8 \times 10^{-3} \, \text{C} = 8 \, \text{mC}

So, the extra charge is

ΔQ=Q2−Q1=8 mC−4 mC=4 mC\Delta Q = Q_2 - Q_1 = 8 \, \text{mC} - 4 \, \text{mC} = 4 \, \text{mC}

Now use the energy formula for constant potential:

U=12CV2U = \frac{1}{2}CV^2

Initial energy:

U1=12×40×10−6×(100)2=0.2 JU_1 = \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2 = 0.2 \, \text{J}

Final energy:

U2=12×80×10−6×(100)2=0.4 JU_2 = \frac{1}{2} \times 80 \times 10^{-6} \times (100)^2 = 0.4 \, \text{J}

Hence, the change in electrostatic energy is

ΔU=U2−U1=0.4−0.2=0.2 J\Delta U = U_2 - U_1 = 0.4 - 0.2 = 0.2 \, \text{J}

Therefore, the extra charge is 4 mC4\,\text{mC} and the increase in electrostatic energy is 0.2 J0.2\,\text{J}. The correct option is C.

Using change formulas directly

Given: C=40×10−6 FC = 40 \times 10^{-6} \, \text{F}, V=100 VV = 100 \, \text{V}, K=2K = 2.

Find: Extra charge and change in electrostatic energy.

The increase in capacitance due to dielectric insertion is

ΔC=C′−C=KC−C=(K−1)C\Delta C = C' - C = KC - C = (K-1)C

Thus,

ΔQ=ΔC⋅V=(KC−C)V\Delta Q = \Delta C \cdot V = (KC - C)V

Substituting the values,

ΔQ=(2×40×10−6−40×10−6)×100\Delta Q = (2 \times 40 \times 10^{-6} - 40 \times 10^{-6}) \times 100 =40×10−6×100= 40 \times 10^{-6} \times 100 =4×10−3 C=4 mC= 4 \times 10^{-3} \, \text{C} = 4 \, \text{mC}

For energy change,

ΔU=12C′V2−12CV2=12(C′−C)V2=12(K−1)CV2\Delta U = \frac{1}{2}C'V^2 - \frac{1}{2}CV^2 = \frac{1}{2}(C' - C)V^2 = \frac{1}{2}(K-1)CV^2

Now substitute the values,

ΔU=12(2−1)(40×10−6)(100)2\Delta U = \frac{1}{2}(2-1)(40 \times 10^{-6})(100)^2 =12×40×10−6×10000= \frac{1}{2} \times 40 \times 10^{-6} \times 10000 =0.2 J= 0.2 \, \text{J}

Therefore, the required values are 4 mC4\,\text{mC} and 0.2 J0.2\,\text{J}.

Common mistakes

  • Using the isolated-capacitor condition instead of the battery-connected condition is incorrect because here the power supply remains connected, so VV stays constant. Use Q=CVQ = CV with constant VV, not constant charge.

  • Taking the final charge Q2=8 mCQ_2 = 8\,\text{mC} as the extra charge is wrong. The question asks for the additional charge supplied, so calculate ΔQ=Q2−Q1\Delta Q = Q_2 - Q_1.

  • Using only the final energy 0.4 J0.4\,\text{J} as the change in energy is incorrect. The change means the increase from the initial state, so use ΔU=U2−U1\Delta U = U_2 - U_1.

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