An electric dipole is placed as shown in the figure. The electric potential (in 102 V) at point P due to the dipole is ( ε₀ = permittivity of free space and (1)/(4πε₀) = K): (a) (5/8)qK (b) (8/5)qK (c) (8/3)qK (d) (3/8)qK

Answer: (d). P is on the axis, 5 cm from O: distance from +q is 2 cm, from -q is 8 cm. V = Kq(1/0.02 - 1/0.08) = Kq(50 - 12.5) = 37.5 Kq 37.5 Kq = 3/8Kq × 10² V So the answer is (d) (3/8)qK.

An electric dipole is placed as shown in the figure. The electric… | NEET 2023

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Electric Dipole

An electric dipole is placed as shown in the figure.

The electric potential (in 102 V) at point P due to the dipole is ( $\epsilon_0$ = permittivity of free space and $\frac{1}{4\pi\epsilon_0}$ = K):

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Answer: (d)

P is on the axis, 5 cm from O: distance from $+q$ is $2$ cm, from $-q$ is $8$ cm.
$V = Kq\left(\frac{1}{0.02} - \frac{1}{0.08}\right) = Kq(50 - 12.5) = 37.5\,Kq$
$37.5\,Kq = \frac{3}{8}Kq \times 10^2$ V
So the answer is (d) $\left(\frac{3}{8}\right)qK$ .

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