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A cube of side 'b' has a charge 'q' at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube. |
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Answer» Solution :`sqrt(b^(2)+2b^(2))/(2)=b/2 sqrt3` is the CENTRE Distance of centre of the cube form one vertex is `b/2 sqrt3`. Hence potential due to charge `=(1)/(4PI epsi_(0)) q/(b/2 sqrt3)` Total potential `=(8 xx q)/(4pi epsi_(0) ""b/2 sqrt3)` `=(2 xx 8 xx q)/(4pi epsi_(0)b sqrt3)=(4q)/(sqrt3 b PI epsi_(0))` Field =zero 
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