Electric Dipole

The electric potential at a point at distance (r) and angle (\theta) from an electric dipole is (V = \frac{kp\cos\theta}{r^2}), where (p) is the dipole moment and (\theta) is the angle with the dipole axis. This result follows from superposing the potentials of the two constituent charges (+q) and (-q) separated by (d), taking the limit (d \to 0) with (qd = p) constant. Here, (p = 2 \times 10^{-9}) C·m, (r = 0.30) m, and (\theta = 60°) so (\cos 60° = 0.5). Substituting: (V = \frac{9 imes 10^9 \times 2 \times 10^{-9} \times 0.5}{(0.30)^2} = \frac{9}{0.09} = 100 \times 0.5 = 25)... Recalculating: (V = 9 imes 10^9 imes 2 imes 10^{-9} imes 0.5 / 0.09 = 9 imes 1 / 0.09 = 100) V... Let me redo: numerator (= 9 \times 10^9 \times 2 \times 10^{-9} \times 0.5 = 9 \times 2 \times 0.5 = 9). Denominator (= (0.3)^2 = 0.09). So (V = 9/0.09 = 100) V. But the (\cos\theta) factor already included: (V = kp\cos\theta/r^2 = 9 \times 10^9 \times 2 \times 10^{-9} \times 0.5 / 0.09 = 9/0.09 = 100) V. Correcting: the correct answer here is 100 V. Option 25 V incorrectly applies the angle twice. Option 50 V uses (\cos 60°) but applies it to the wrong expression. Option 12.5 V applies an additional factor incorrectly. The dipole potential formula is a standard JEE Advanced topic. Plausibility check: 100 V at 30 cm from a nanocoulomb-metre dipole is consistent with typical JEE problem scales.