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If the force exerted by an electric dipole on a charge $\mathrm{q}$ at a distance of $1 \mathrm{~m}$ is $\mathrm{F}$, the force at a point $2 \mathrm{~m}$ away in the same direction will be
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The correct answer is:
$\frac{\mathrm{F}}{8}$
We know that the electric field at any point due to an electric dipole varies inversely with the cube of the distance of the point from the centre of the dipole, that is,
$\mathrm{E} \propto \frac{1}{r^{3}}$
Also, force on a charge (q) is given by
$\mathrm{F}=\mathrm{qE} \Rightarrow \mathrm{F} \propto \frac{1}{\mathrm{r}^{3}}$
When the distance of the charge becomes $2 \mathrm{~m}$, i.e. double of its initial value, then new force $\left(\mathrm{F}^{\prime}\right)$ will become
$\mathrm{F}^{\prime}=\frac{1}{(2)^{3}} \cdot \mathrm{F}=\frac{\mathrm{F}}{8}$
$\mathrm{E} \propto \frac{1}{r^{3}}$
Also, force on a charge (q) is given by
$\mathrm{F}=\mathrm{qE} \Rightarrow \mathrm{F} \propto \frac{1}{\mathrm{r}^{3}}$
When the distance of the charge becomes $2 \mathrm{~m}$, i.e. double of its initial value, then new force $\left(\mathrm{F}^{\prime}\right)$ will become
$\mathrm{F}^{\prime}=\frac{1}{(2)^{3}} \cdot \mathrm{F}=\frac{\mathrm{F}}{8}$
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