Six charges +q, -q, +q, -q, +q and -q are fix | vedclass

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(A) The work done in bringing a charge $q_0$ from infinity to a point is given by $W = q_0 V$,where $V$ is the electric potential at that point due to

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$0$

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(A) The work done in bringing a charge $q_0$ from infinity to a point is given by $W = q_0 V$,where $V$ is the electric potential at that point due to the existing charges.In a regular hexagon of side $d$,the distance of each corner from the centre is $d$.The electric potential $V$ at the centre due to the six charges is the sum of the potentials due to each individual charge:$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{+q}{d} + \frac{-q}{d} + \frac{+q}{d} + \frac{-q}{d} + \frac{+q}{d} + \frac{-q}{d} ight)$$V = \frac{1}{4 \pi \varepsilon_0 d} (q - q + q - q + q - q)$$V = 0$Therefore,the work done $W = q_0 	imes 0 = 0$.

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 $q_0$ to the centre of the hexagon from infinity is: ($\varepsilon_0$ = permittivity of free space)

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