Two electric dipoles of moment $\mathrm{P}$ and $27 \mathrm{P}$ are placed on a line with their centres $24…
- $6 \mathrm{~cm}$
- $8 \mathrm{~cm}$
- $10 \mathrm{~cm}$
- $12 \mathrm{~cm}$
Solution
At N,
$\mid \mathrm{E} . \mathrm{F}$. due to dipole $1|=| \mathrm{E}$. F. due to dipole $2 \mid$
$\begin{aligned}
& \therefore \quad \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 \mathrm{p}}{\mathrm{x}^3}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2(27 \mathrm{p})}{(24-\mathrm{x})^3} \\
& \therefore \quad \frac{1}{\mathrm{x}^3}=\frac{27}{(24-\mathrm{x})^3} \Rightarrow \mathrm{x}=6 \mathrm{~cm} .
\end{aligned}$Asked in: MHT CET 2023 (12 May Shift 1)