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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?
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Verified Answer
The correct answer is:
$6 \mathrm{~cm}$
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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