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Question

Six charges +q, $-$q, +q, $-$q, +q, and $-$q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q0 to the centre of the hexagon from infinity is

(${\varepsilon _0}$ - permittivity of free space)

${{ - {q^2}} \over {4\pi {\varepsilon _0}d}}\left( {6 - {1 \over {\sqrt 2 }}} \right)$
Zero
${{ - {q^2}} \over {4\pi {\varepsilon _0}d}}$
${{ - {q^2}} \over {4\pi {\varepsilon _0}d}}\left( {3 - {1 \over {\sqrt 2 }}} \right)$

Solution

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