MHT CET · Physics · Magnetic Effects of Current
Two electric dipoles of moment \(\mathrm{P}\) and \(27 \mathrm{P}\) are placed on a line with their centres \(24 \mathrm{~cm}\) apart. Their dipole moments are in opposite direction. At which point the electric field will be zero between the dipoles from the centre of dipole of moment P?
- A \(6 \mathrm{~cm}\)
- B \(8 \mathrm{~cm}\)
- C \(10 \mathrm{~cm}\)
- D \(12 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(6 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Dipole of moment \(\mathrm{p}\) is at a distance \(\mathrm{x}\) from \(\mathrm{N}\)

At N,
\(\mid \mathrm{E} . \mathrm{F}\). due to dipole \(1|=| \mathrm{E}\). F. due to dipole \(2 \mid\)
\(
\begin{aligned}
& \therefore \quad \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 \mathrm{p}}{\mathrm{x}^3}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2(27 \mathrm{p})}{(24-\mathrm{x})^3} \\
& \therefore \quad \frac{1}{\mathrm{x}^3}=\frac{27}{(24-\mathrm{x})^3} \Rightarrow \mathrm{x}=6 \mathrm{~cm} .
\end{aligned}
\)

At N,
\(\mid \mathrm{E} . \mathrm{F}\). due to dipole \(1|=| \mathrm{E}\). F. due to dipole \(2 \mid\)
\(
\begin{aligned}
& \therefore \quad \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 \mathrm{p}}{\mathrm{x}^3}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2(27 \mathrm{p})}{(24-\mathrm{x})^3} \\
& \therefore \quad \frac{1}{\mathrm{x}^3}=\frac{27}{(24-\mathrm{x})^3} \Rightarrow \mathrm{x}=6 \mathrm{~cm} .
\end{aligned}
\)
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