Electric Potential: Practice Problem & Solution
A short electric dipole has a dipole moment of $16 \times 10^{-9}\text{ C m}$. The electric potential due to the dipole at a point at a distance of $0.6\text{ m}$ from the centre of the dipole, situated on a line making an angle of $60^{\circ}$ with the dipole axis is : $\left(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N m}^2/\text{C}^2\right)$ (2020)
Solution Explained:
To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:
Electric potential due to a short dipole is $V = \frac{kp\cos\theta}{r^2}$. Substituting the values: $V = \frac{9 \times 10^9 \times 16 \times 10^{-9} \times \cos(60^{\circ})}{(0.6)^2} = \frac{144 \times 0.5}{0.36} = 200\text{ V}$.
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