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 and E 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: Calculate the electric potential at point \( B \).The electric potential from a dipole at any location is defined as:\[V = \frac{1}{4\pi\epsilon_0} \frac{\mathbf{p} \cdot \hat{r}}{r^2}.\]As point \( B \) lies on the perpendicular bisector of the dipole, \( \mathbf{p} \cdot \hat{r} = 0 \). Therefore:\[V_B = 0.\]Step 2: Calculate the electric field at point \( B \).The electric field's magnitude on the perpendicular bisector of a dipole is given by:\[E = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + d^2)^{3/2}}.\]Applying the small dipole approximation \( d \ll r \), we get:\[E_B = \frac{E_0}{16}.\]Consequently, the result is \( \boxed{0, \frac{E_0}{16}} \).