Question 41: An electric dipole is placed as shown in the figure. The electric potential (in 10^{2} \mathrm{~V} ) at point P due to the dipole is ( \epsilon_{0}= permittivity of free space and \frac{1}{4 \pi \epsilon_{0}}=\mathrm{K} ):
(1) \left(\frac{8}{5}\right) \mathrm{qK}
(2) \left(\frac{8}{3}\right) \mathrm{qK}
(3) \left(\frac{3}{8}\right) \mathrm{qK}
(4) \left(\frac{5}{8}\right) \mathrm{qK}
Answer: Option (3)
Explanation:
Electric potential due to a point charge is given by V = \frac{Kq}{r}.
The net potential due to a dipole is the algebraic sum of potentials due to individual charges.
From the figure, the distance between the charges is 6 \mathrm{~cm}.
The midpoint O is 3 \mathrm{~cm} away from each charge.
Point P is located 5 \mathrm{~cm} to the right of the midpoint O.
Hence, distance of point P from charge
+q is r_{+} = 5 - 3 = 2 \mathrm{~cm}
and from charge -q is r_{-} = 5 + 3 = 8 \mathrm{~cm}.
The potential at P due to +q is V_{+} = \frac{Kq}{2}
and due to -q is V_{-} = -\frac{Kq}{8}.
The net potential at P is
V = \frac{Kq}{2} - \frac{Kq}{8}.
Taking LCM,
V = \frac{4Kq - Kq}{8} = \frac{3Kq}{8}.
Thus, the electric potential at point P is \left(\frac{3}{8}\right) \mathrm{qK} in units of 10^{2} \mathrm{~V}.
Hence, the correct answer is Option (3).