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Q.

Two dipoles each of moment 5 × 10^–12 Cm form a cross with their axis (-to+) along the coordinate axes. The potential at a point 20cm away in a direction making an angle of $30^{o}$ with axis is

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a

1.12 V

b

2.4 V

c

2.12 V

d

1.536 V

answer is D.

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Detailed Solution

$\begin{array}{l} V = V_{1} + V_{2} \\ = \frac{1}{4 {πε}_{0}} \frac{Pcos 30^{0}}{r^{2}} + \frac{1}{4 {πε}_{0}} \frac{Pcos (90 - 30^{0})}{r^{2}} \\ = \frac{P}{4 {πε}_{0} r^{2}} [\frac{\sqrt{3}}{2} + \frac{1}{2}] = \frac{9 \times 10^{9} \times 5 \times 10^{- 12} \times 2 .732}{0 .2 \times 0 .2 \times 2} \end{array}$
=1.536V

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