JEE MAIN - Physics (2024 - 5th April Evening Shift - No. 21)
The electric field at point $$\mathrm{p}$$ due to an electric dipole is $$\mathrm{E}$$. The electric field at point $$\mathrm{R}$$ on equitorial line will be $$\frac{\mathrm{E}}{x}$$. The value of $$x$$ :
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Answer
16
Explanation
$$\begin{aligned}
& E=\frac{2 k p}{r^3} \\
& E_R=\frac{k p}{(2 r)^3}=\frac{1}{8}\left(\frac{E}{2}\right) \\
& =\frac{E}{16} \\
& \therefore \quad x=16 \\
&
\end{aligned}$$
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