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Capacitor question

2003 · Shift 0 · Q169
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Capacitor question

2003 · Shift 0 · Q169

JEE MainPhysicsCapacitorMCQ+4 / −1
The work done in placing a charge of 8×10−188 \times {10^{ - 18}}8×10−18 coulomb on a condenser of capacity 100100100 micro-farad is
  1. A
    16×10−32  joule16 \times {10^{ - 32}}\,\,joule16×10−32joule
  2. B
    3.1×10−26  joule3.1 \times {10^{ - 26}}\,\,joule3.1×10−26joule
  3. C
    4×10−10  joule4 \times {10^{ - 10}}\,\,joule4×10−10joule
  4. D
    32×10−32  joule32 \times {10^{ - 32}}\,\,joule32×10−32joule
View written solutionFree

Correct answer: D

  1. Use the energy stored in a capacitor

The work done in placing charge qqq on a capacitor of capacitance CCC is equal to the electrostatic energy stored:

W=q22CW = \frac{q^2}{2C}W=2Cq2​

  1. Given data

q=8×10−18 Cq = 8 \times 10^{-18}\ \text{C}q=8×10−18 C C=100 μF=100×10−6 F=10−4 FC = 100\ \mu \text{F} = 100 \times 10^{-6}\ \text{F} = 10^{-4}\ \text{F}C=100 μF=100×10−6 F=10−4 F

  1. Substitute into the formula

W=(8×10−18)22×10−4W = \frac{(8 \times 10^{-18})^2}{2 \times 10^{-4}}W=2×10−4(8×10−18)2​

First, square the charge:

(8×10−18)2=64×10−36(8 \times 10^{-18})^2 = 64 \times 10^{-36}(8×10−18)2=64×10−36

So,

W=64×10−362×10−4W = \frac{64 \times 10^{-36}}{2 \times 10^{-4}}W=2×10−464×10−36​

W=32×10−32 JW = 32 \times 10^{-32}\ \text{J}W=32×10−32 J

  1. Match with the options

32×10−32 J32 \times 10^{-32}\ \text{J}32×10−32 J

This matches Option D.

  1. Answer check

Stored correct answer: D

Our derived answer: D

So, the stored answer is correct.

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