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A uniform electric field 400 N/C acts along positive y-axis. P is a point having co-ordinates (0.6 m, -0.2 m) and R another point, with co-ordinates (-0.4 m, 0.6 m). If electric potential at P is 200 V, potential at R will be :

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100 V

-150 V

80 V

-120 V

answer is D.

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Detailed Solution

$∆ V = V_{R} - V_{P} = - \vec{E} . ∆ \vec{r} = - \vec{E} (\vec{r_{R}} - \vec{r_{P}}) \Rightarrow V_{R} - 200 = - (- 400 \hat{j}) . (- \hat{i} + 0.8 \hat{j}) \Rightarrow V_{R} - 200 = - 320 \Rightarrow V_{R} = - 120 volt$

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