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In a uniform electric field the potential at origin is V and V/2 at points (a, 0, 0), (0, b, 0) and (0, 0, c). Then potential at (a, b, c) will be

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a

– V

b

$- \frac{3 V}{2}$

c

$\frac{V}{2}$

d

$- \frac{V}{2}$

answer is C.

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Detailed Solution

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$Let the electric field be \vec{E} = E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k} ∴ at (a, 0, 0), \frac{V}{2} - V = - E_{x} a \Rightarrow E_{x} = \frac{V}{2 a} Similarly, E_{y} = \frac{V}{2 b} and E_{z} = \frac{V}{2 c} ∴ at (a, b, c), V' - V = - E_{x} . a - E_{y} . b - E_{z} . c = - \frac{3 V}{2} \Rightarrow V' = - \frac{V}{2}$

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