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Electrostatics question

2009 · Shift 0 · Q59
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Electrostatics question

2009 · Shift 0 · Q59

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Two points PPP and QQQ are maintained at the potentials of 10V10V10V and −4V-4V−4V, respectively. The work done in moving 100100100 electrons from PPP to QQQ is :
  1. A
    9.60×10−17J9.60 \times {10^{ - 17}}J9.60×10−17J
  2. B
    −2.24×10−16J- 2.24 \times {10^{ - 16}}J−2.24×10−16J
  3. C
    2.24×10−16J2.24 \times {10^{ - 16}}J2.24×10−16J
  4. D
    −9.60×10−17J- 9.60 \times {10^{ - 17}}J−9.60×10−17J
View written solutionFree

Correct answer: C

  1. Given data

    • Potential at PPP: VP=10 VV_P = 10\,\text{V}VP​=10V
    • Potential at QQQ: VQ=−4 VV_Q = -4\,\text{V}VQ​=−4V
    • Number of electrons: n=100n = 100n=100
    • Charge of one electron: e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C

  2. Potential difference from PPP to QQQ

    ΔV=VQ−VP=(−4)−10=−14 V\Delta V = V_Q - V_P = (-4) - 10 = -14\,\text{V}ΔV=VQ​−VP​=(−4)−10=−14V

  3. Total charge of 100 electrons Since each electron has charge −e-e−e,

    q=−100×1.6×10−19q = -100 \times 1.6 \times 10^{-19}q=−100×1.6×10−19 q=−1.6×10−17 Cq = -1.6 \times 10^{-17}\,\text{C}q=−1.6×10−17C

  4. Work done Work done in moving charge qqq through potential difference ΔV\Delta VΔV is

    W=qΔVW = q\Delta VW=qΔV

    Substituting,

    W=(−1.6×10−17)(−14)W = (-1.6 \times 10^{-17})(-14)W=(−1.6×10−17)(−14) W=2.24×10−16 JW = 2.24 \times 10^{-16}\,\text{J}W=2.24×10−16J

  5. Matching with options

    2.24×10−16 J2.24 \times 10^{-16}\,\text{J}2.24×10−16J

    This corresponds to Option C.

  6. Final answer The work done in moving 100100100 electrons from PPP to QQQ is

    2.24×10−16 J\boxed{2.24 \times 10^{-16}\,\text{J}}2.24×10−16J​

    Hence, the correct option is C.

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