Question 606: Electrostatics & Capacitance

Question

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 $$x$$ :

Accepted Answer

$$\begin{aligned}
& E=\frac{2 k p}{r^3} \
& E_R=\frac{k p}{(2 r)^3}=\frac{1}{8}\left(\frac{E}{2}ight) \
& =\frac{E}{16} \
& 	herefore \quad x=16 \
&
\end{aligned}$$

Electrostatics & Capacitance

The electric field at point $\mathrm{p} due to an electric dipole is \mathrm{E}. The electric field at point \mathrm{R} on equitorial line will be \frac{\mathrm{E}}{x}. The value of x$ :

Integer / Numeric Answer Type