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Electrostatics question

2007 · Q147
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Electrostatics question

2007 · Q147

NEETPhysicsElectrostaticsMCQ+4 / −1
Charges +q and −-−q are placed at points A and B respectively which are a distance 2L apart, C is the midnight between A and B. The work done in moving a charge + Q along the semicircle CRD is

AIPMT 2007 Physics - Electrostatics Question 43 English
  1. A
    qQ2πε0L{{qQ} \over {2\pi {\varepsilon _0}L}}2πε0​LqQ​
  2. B
    qQ6πε0L{{qQ} \over {6\pi {\varepsilon _0}L}}6πε0​LqQ​
  3. C
    −-− qQ6πε0L{{qQ} \over {6\pi {\varepsilon _0}L}}6πε0​LqQ​
  4. D
    qQ4πε0L{{qQ} \over {4\pi {\varepsilon _0}L}}4πε0​LqQ​
View written solutionFree

Correct answer: C

AIPMT 2007 Physics - Electrostatics Question 43 English Explanation

Potential at C = VC = 0
Potential at D = VD

=k(−qL)+kq3L=−23kqL = k\left( {{{ - q} \over L}} \right) + {{kq} \over {3L}} = - {2 \over 3}{{kq} \over L}=k(L−q​)+3Lkq​=−32​Lkq​

Potential difference
VD – VC =−23kqL=14πε0(−23.qL) = - {2 \over 3}{{kq} \over L} = {1 \over {4\pi {\varepsilon _0}}}\left( { - {2 \over 3}.{q \over L}} \right)=−32​Lkq​=4πε0​1​(−32​.Lq​)

⇒\Rightarrow⇒ Work done = Q (VD – VC)

=−23×14πε0qQL=−qQ6πε0L = - {2 \over 3} \times {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over L} = {{ - qQ} \over {6\pi {\varepsilon _0}L}}=−32​×4πε0​1​LqQ​=6πε0​L−qQ​

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