Question

In broad-side-on position, the electric potential due to an electric dipole is:

• A. \( \frac{p}{4\pi \varepsilon_0 r} \)
• B. \( \frac{-p}{4\pi \varepsilon_0 r^2} \)
• C. zero
• D. infinite

Hint:

Electric potential due to a dipole: \[ V = \frac{1}{4\pi \varepsilon_0} \cdot \frac{\vec{p} \cdot \hat{r}}{r^2} \] - Axial line (end-on): \( V = \frac{p}{4\pi \varepsilon_0 r^2} \) - Equatorial line (broad-side-on): \( V = \frac{-p}{4\pi \varepsilon_0 r^2} \)

Answer 1

The Correct Option is B

The electric potential \( V \) due to an electric dipole at a point depends on the angle \( \theta \) between the dipole axis and the position vector. In the broad-side-on position, the point lies on the equatorial line of the dipole (i.e., perpendicular to the dipole axis), so \( \theta = 90^\circ \). The potential at a point on the equatorial line is given by: \[ V = \frac{1}{4\pi \varepsilon_0} \cdot \frac{p \cos\theta}{r^2} \] Since \( \cos(90^\circ) = 0 \), this would seem to give \( V = 0 \), but this formula is valid for general angles. On the equatorial line (broad-side-on), the correct potential is: \[ V = \frac{-p}{4\pi \varepsilon_0 r^2} \]