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Electrostatics question

2011 · Q163
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Electrostatics question

2011 · Q163

NEETPhysicsElectrostaticsMCQ+4 / −1
Four electric charges +q, +q, −-− q and −-− q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, midway between the two charges + q and +q, is

AIPMT 2011 Prelims Physics - Electrostatics Question 54 English
  1. A
    14πε02qL(1+5){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 + \sqrt 5 } \right)4πε0​1​L2q​(1+5​)
  2. B
    14πε02qL(1+15){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 + {1 \over {\sqrt 5 }}} \right)4πε0​1​L2q​(1+5​1​)
  3. C
    14πε02qL(1−15){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 - {1 \over {\sqrt 5 }}} \right)4πε0​1​L2q​(1−5​1​)
  4. D
    zero
View written solutionFree

Correct answer: C

Distance of point A from the two +q charges = L.

Distance of point A from the two –q charges
=L2+(2L)2=5L = \sqrt {{L^2} + {{\left( {2L} \right)}^2}} = \sqrt 5 L=L2+(2L)2​=5​L

AIPMT 2011 Prelims Physics - Electrostatics Question 54 English Explanation


∴\therefore∴ VA=(KqL×2)−(Kq5L×2){V_A} = \left( {{{Kq} \over L} \times 2} \right) - \left( {{{Kq} \over {\sqrt 5 L}} \times 2} \right)VA​=(LKq​×2)−(5​LKq​×2)

=2KqL[1−15] = {{2Kq} \over L}\left[ {1 - {1 \over {\sqrt 5 }}} \right]=L2Kq​[1−5​1​]

=14πε0.2qL(1−15) = {1 \over {4\pi {\varepsilon _0}}}.{{2q} \over L}\left( {1 - {1 \over {\sqrt 5 }}} \right)=4πε0​1​.L2q​(1−5​1​)

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