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Q.

An electric dipole is formed by two charges +q and –q located in xy-plane at (0, 2) mm and (0, −2) mm, respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is V₀. The charges +q and –q are then moved to the points (−1, 2) mm and (1, −2) mm, respectively. What is the value of electric potential at P due to the new dipole?

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a

$\frac{V_{0}}{\sqrt{2}}$

b

$\frac{3 V_{0}}{4}$

c

$\frac{V_{0}}{4}$

d

$\frac{V_{0}}{2}$

answer is B.

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Detailed Solution

$\frac{Kpcos θ}{r^{2}} = V_{0} (\cos θ = \frac{1}{\sqrt{2}}) \frac{KP}{r^{2}} = \sqrt{2} V_{0}$
Now new dipole momentum $P^{'} = \frac{p}{A_{2}} \times λ \sqrt{5}$
Hence $V^{'} = \frac{{KP}^{'} \cos (45 + θ)}{r^{2}} = \frac{KP \sqrt{5}}{2 r^{2}} (\cos 45 \cdot \cos θ - \sin 45 \cdot \sin θ)$
$= \sqrt{2} V_{0} \frac{\sqrt{5}}{2} (\frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{5}}) = 2 \frac{V_{0}}{2} \frac{2 - 1}{\sqrt{2}} = \frac{V_{0}}{2}$

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An electric dipole is formed by two charges +q and –q located in xy-plane at (0, 2) mm and (0, −2) mm, respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is V0. The charges +q and –q are then moved to the points (−1, 2) mm and (1, −2) mm, respectively. What is the value of electric potential at P due to the new dipole?