For a short dipole placed at origin O, the dipole moment P is along 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

Understanding the Problem

We have a short dipole placed at the origin $O$ with its dipole moment $P$ along the x-axis.

• Point A is on the axial line at a distance $r$ from the dipole.
• Point B is on the equatorial line at a distance $2r$ from the dipole.

Given:

• The electric potential at $A$ is $V_0$.
• The electric field at $A$ is $E_0$.

We need to determine the electric potential and electric field at point B.

Step 1: Electric Potential of a Dipole

The electric potential $V$ due to a short dipole at a point at distance $r$ is:

$$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{\mathbf{p} \cdot \hat{r}}{r^2}$$

On the axial line (Point A at distance $r$):

• $$V_A = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^2} = V_0$$

On the equatorial line (Point B at distance $2r$):

Since the dipole potential is given by:

• $$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{p \cos\theta}{r^2}$$

and on the equatorial line $\theta = 90^\circ \Rightarrow \cos90^\circ = 0$,

• $$V_B = 0$$

Thus, the electric potential at B is zero.

Step 2: Electric Field of a Dipole

The magnitude of the electric field at a distance $r$ from a dipole:

On the axial line:

$$E_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}$$

Given that at $A$ (on the axial line at $r$), the field is $E_0$:

$$E_A = E_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}$$

On the equatorial line:

$$E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3}$$

At B (distance $2r$):

$$E_B = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{(2r)^3} = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{8r^3}$$

Since $E_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}$, we can express $E_B$ in terms of $E_0$:

$$E_B = \frac{E_0}{16}$$

Final Answer:

• Electric potential at B = $0$
• Electric field at B = $\frac{E_0}{16}$