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Electrostatics question

2023 · 29 Jan · Shift 2 · Q61
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Electrostatics question

2023 · 29 Jan · Shift 2 · Q61

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A point charge 2×10−2 C2\times10^{-2}~\mathrm{C}2×10−2 C is moved from P to S in a uniform electric field of 30 NC−130~\mathrm{NC^{-1}}30 NC−1 directed along positive x-axis. If coordinates of P and S are (1, 2, 0) m and (0, 0, 0) m respectively, the work done by electric field will be
  1. A
    600 mJ
  2. B
    −1200-1200−1200 mJ
  3. C
    1200 mJ
  4. D
    −600-600−600 mJ
View written solutionFree

Correct answer: D

  1. Given data
  • Charge: q=2×10−2 Cq = 2\times 10^{-2}\,\mathrm{C}q=2×10−2C
  • Uniform electric field: E⃗=30 N/C\vec E = 30\,\mathrm{N/C}E=30N/C along positive xxx-axis
  • Initial point: P(1,2,0)P(1,2,0)P(1,2,0)
  • Final point: S(0,0,0)S(0,0,0)S(0,0,0)

So,

E⃗=30i^\vec E = 30\hat iE=30i^
  1. Displacement from PPP to SSS

The displacement vector is

Δr⃗=r⃗S−r⃗P=(0−1)i^+(0−2)j^+(0−0)k^\Delta \vec r = \vec r_S - \vec r_P = (0-1)\hat i + (0-2)\hat j + (0-0)\hat kΔr=rS​−rP​=(0−1)i^+(0−2)j^​+(0−0)k^ Δr⃗=−i^−2j^\Delta \vec r = -\hat i - 2\hat jΔr=−i^−2j^​
  1. Work done by electric field

For a uniform electric field, work done by the field is

W=q E⃗⋅Δr⃗W = q\,\vec E\cdot \Delta \vec rW=qE⋅Δr

Now,

E⃗⋅Δr⃗=(30i^)⋅(−i^−2j^)=30(−1)+0=−30\vec E\cdot \Delta \vec r = (30\hat i)\cdot(-\hat i-2\hat j)=30(-1)+0=-30E⋅Δr=(30i^)⋅(−i^−2j^​)=30(−1)+0=−30

Therefore,

W=(2×10−2)(−30)W = \left(2\times 10^{-2}\right)(-30)W=(2×10−2)(−30) W=−0.6 JW = -0.6\,\mathrm{J}W=−0.6J
  1. Convert into mJ

Since

1 J=1000 mJ1\,\mathrm{J} = 1000\,\mathrm{mJ}1J=1000mJ

we get

−0.6 J=−600 mJ-0.6\,\mathrm{J} = -600\,\mathrm{mJ}−0.6J=−600mJ
  1. Option check
  • A: 600 mJ600\,\mathrm{mJ}600mJ
  • B: −1200 mJ-1200\,\mathrm{mJ}−1200mJ
  • C: 1200 mJ1200\,\mathrm{mJ}1200mJ
  • D: −600 mJ-600\,\mathrm{mJ}−600mJ

Hence, the correct option is D.

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