Explanation

(d) Let $AB$ be a dipole of length$\ l$. Hence $O$ be the midpoint of the dipole.
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Then the electric field at point $O$, due to $+ q$ charge
$E_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{\left( \frac{l}{2} \right)^{2}}$
$\Longrightarrow E_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{4q}{l^{2}}\ \ \ \ \ \ \ \ (along\ OA)$
and the electric field at point $O,$ due to –q charge
$E_{2} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{\left( \frac{l}{2} \right)^{2}}$
$\Longrightarrow E_{2} = \frac{1}{4\pi\varepsilon_{0}}\frac{4q}{l^{2}}\ \ \ \ \ \ \ \ (along\ OA)$
Hence, the total electric field at point $O$
$E = E_{1} + E_{2}$
$E = 2.\frac{1}{4\pi\varepsilon_{0}}\frac{4q}{l^{2}}$
$\Longrightarrow \ \ E = \frac{1}{4\pi\varepsilon_{0}}\frac{8q}{l^{2}}\ \ \ \ \ldots.(i)$
Now, potential at $O$, due to $+ q$ charge
$V_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{l/2}$
$\Longrightarrow V_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{2q}{l}$
and potential at point $O$, due to $–q$ charge
$V_{2} = \frac{1}{4\pi\varepsilon_{0}}\ \frac{- q}{l/2}$
$\Longrightarrow \ \ V_{2} = - \frac{1}{4\pi\varepsilon_{0}}\frac{2q}{l}$
Hence, total potential at point $O$,
$V = 0$ ………………….(ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{E}{V} = \infty$