A uniform electric field of 100 Vm^-1 is directed at 60^o with the positive x-axis as shown in figure. If OA = 2m, the potential difference V_O - V_A is

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100 V

50 V

-50 V

-100 V

answer is D.

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Detailed Solution

$\begin{array}{l} OB = OAcos 60^{\circ} = 2 \times \frac{1}{2} = 1 m \\ V_{B} - V_{O} = E (OB) = 100 \times 1 = 100 V \\ V_{A} - V_{O} = 100 V [∵ V_{B} = V_{A}] \\ V_{O} - V_{A} = - 100 V \end{array}$

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