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
B
4Q4πε0a
C
4Q3πε0a
D
2Qπε0a

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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