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Question

A charge $$‘Q’$$ is placed at each corner of a cube of side $$‘a’$$. The potential at the centre of the cube is :



A
8Qπε0a
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B
4Q4πε0a
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C
4Q3πε0a
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D
2Qπε0a
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Solution

The correct option is D $$\dfrac{4Q}{\sqrt{3}\pi \varepsilon _{0}a}$$
Length of body diagonal  $$ =\sqrt{3}a . $$
$$\therefore$$ distance between charge Q and centre  $$ =\dfrac{\sqrt{3}a}{2}$$
We know $$V =\dfrac{kQ}{r}$$
As potential is scalar quantity $$\Rightarrow$$ potential due to all charges at the centre add up  
$$ \Rightarrow V= 8\times \dfrac{k\times Q}{\left ( \dfrac{\sqrt{3}a}{{2}} \right )}$$
$$=\dfrac{4Q}{\sqrt{3}\pi  \epsilon_o a}$$

Physics

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