1. Potential due to a short dipole

For a short dipole of moment $\vec p$ at distance $r$, the potential at a point making angle $\theta$ with the dipole axis is

$V=\frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$

2. Electric field due to a short dipole

Magnitude of electric field:

• On the axial line ($\theta=0$):

$E_{\text{axial}}=\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$

• On the equatorial line ($\theta=90^\circ$):

$E_{\text{equatorial}}=\frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}$

3. Identify point A

Since $\vec p$ is along the $x$-axis, point $A$ lies on the dipole axis in the figure, and point $B$ lies on the $y$-axis, i.e. on the equatorial line.

Let distance of $A$ from origin be $r$. From the figure, point $B$ is at twice the distance from origin, i.e.

$OB=2r$

At $A$, the potential is $V_0$ and electric field is $E_0$.

Since $A$ is on axial line,

$V_0=\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}$

and

$E_0=\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$

4. Potential at B

Point $B$ is on the equatorial line, so $\theta=90^\circ$. Thus,

$V_B=\frac{1}{4\pi\varepsilon_0}\frac{p\cos 90^\circ}{(2r)^2}=0$

5. Electric field at B

At equatorial point $B$:

$E_B=\frac{1}{4\pi\varepsilon_0}\frac{p}{(2r)^3}$

$E_B=\frac{1}{4\pi\varepsilon_0}\frac{p}{8r^3}$

Now compare with $E_0$:

$E_0=\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$

Therefore,

$\frac{E_B}{E_0}=\frac{\frac{p}{8r^3}}{\frac{2p}{r^3}}=\frac{1}{16}$

So,

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

6. Final answer

At point $B$:

• Electric potential $=0$
• Electric field $=\dfrac{E_0}{16}$

So the correct option is

$\boxed{\text{D}}$

7. Comparison with stored answer

Stored correct answer: $\text{D}$

Our derived answer also gives $\text{D}$, so they agree.