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Capacitor question

2018 · Shift 0 · Q54
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Capacitor question

2018 · Shift 0 · Q54

JEE MainPhysicsCapacitorMCQ+4 / −1
A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :
  1. A
    0.9 n C
  2. B
    1.2 n C
  3. C
    0.3 n C
  4. D
    2.4 n C
View written solutionFree

Correct answer: B

  1. Given data

    • Initial capacitance: C0=90 pF=90×10−12 FC_0 = 90\,\text{pF} = 90 \times 10^{-12}\,\text{F}C0​=90pF=90×10−12F
    • Battery voltage: V=20 VV = 20\,\text{V}V=20V
    • Dielectric constant: K=53K = \dfrac{5}{3}K=35​

  2. Initial charge on the capacitor Since the capacitor is connected to a battery, the potential difference remains constant.

    Q0=C0V=90×10−12×20=1800×10−12 C=1.8×10−9 CQ_0 = C_0 V = 90 \times 10^{-12} \times 20 = 1800 \times 10^{-12}\,\text{C} = 1.8 \times 10^{-9}\,\text{C}Q0​=C0​V=90×10−12×20=1800×10−12C=1.8×10−9C

    So, Q0=1.8 nCQ_0 = 1.8\,\text{nC}Q0​=1.8nC

  3. Capacitance after inserting dielectric C=KC0=53×90=150 pFC = K C_0 = \frac{5}{3} \times 90 = 150\,\text{pF}C=KC0​=35​×90=150pF

  4. New charge on the capacitor Q=CV=150×10−12×20=3000×10−12 C=3.0 nCQ = CV = 150 \times 10^{-12} \times 20 = 3000 \times 10^{-12}\,\text{C} = 3.0\,\text{nC}Q=CV=150×10−12×20=3000×10−12C=3.0nC

  5. Induced charge due to dielectric The induced charge on the dielectric surface is

    Q′=Q−Q0Q' = Q - Q_0Q′=Q−Q0​

    Q′=3.0−1.8=1.2 nCQ' = 3.0 - 1.8 = 1.2\,\text{nC}Q′=3.0−1.8=1.2nC

    Equivalently, Q′=(K−1)C0V=(53−1)(90 pF)(20 V)=23×1.8 nC=1.2 nCQ' = (K-1)C_0V = \left(\frac{5}{3}-1\right)(90\,\text{pF})(20\,\text{V}) = \frac{2}{3}\times 1.8\,\text{nC} = 1.2\,\text{nC}Q′=(K−1)C0​V=(35​−1)(90pF)(20V)=32​×1.8nC=1.2nC

  6. Option check

    • A: 0.9 nC0.9\,\text{nC}0.9nC ❌
    • B: 1.2 nC1.2\,\text{nC}1.2nC ✅
    • C: 0.3 nC0.3\,\text{nC}0.3nC ❌
    • D: 2.4 nC2.4\,\text{nC}2.4nC ❌

Therefore, the correct option is B.

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