Question

For a short dipole placed at origin \( O \), the dipole moment \( P \) is along the \( x \)-axis, as shown in the figure. If the electric potential and electric field at \( A \) are \( V_0 \) and \( E_0 \) respectively, then the correct combination of the electric potential and electric field, respectively, at point \( B \) on the \( y \)-axis is given by: 

• A. \( \frac{V_0}{4}, \frac{E_0}{4} \)
• B. \( 0, \frac{E_0}{16} \)
• C. \( \frac{V_0}{2}, \frac{E_0}{16} \)
• D. \( \frac{E_0}{8} \)

Hint:

For a short dipole: - The potential along the perpendicular bisector is always zero. - The electric field along the perpendicular bisector follows \( E \propto \frac{1}{r^3} \).

Answer 1

The Correct Option is B

Step 1: Compute the electric potential at point \( B \). The electric potential due to a dipole at any point is given by: \[ V = \frac{1}{4\pi\epsilon_0} \frac{\mathbf{p} \cdot \hat{r}}{r^2}. \] Since point \( B \) is on the perpendicular bisector of the dipole, \( \mathbf{p} \cdot \hat{r} = 0 \), implying: \[ V_B = 0. \] Step 2: Compute the electric field at point \( B \). The magnitude of the electric field along the perpendicular bisector of a dipole is: \[ E = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + d^2)^{3/2}}. \] For small dipole approximation \( d \ll r \), we use: \[ E_B = \frac{E_0}{16}. \] Thus, the answer is \( \boxed{0, \frac{E_0}{16}} \).