Two opposite charges each of magnitude $500 \mu \mathrm{C}$ are $10 \mathrm{~cm}$ apart. Find electric field…
Two opposite charges each of magnitude $500 \mu \mathrm{C}$ are $10 \mathrm{~cm}$ apart. Find electric field intensity at a distance of $25 \mathrm{~cm}$ from the midpoint on axial line of the dipole.
$5.76 \times 10^7 \mathrm{NC}^{-1}$
$9.28 \times 10^7 \mathrm{NC}^{-1}$
$13.1 \times 10^{10} \mathrm{NC}^{-1}$
$20.5 \times 10^7 \mathrm{NC}^{-1}$
Solution
Given, charge of dipole, $q=500 \times 10^{-6} \mathrm{C}$
Separation between charges, $2 a=10 \mathrm{~cm}$
$=10 \times 10^{-2} \mathrm{~m}$
Distance of location from mid point of axis,
$r=25 \mathrm{~cm}=25 \times 10^{-2} \mathrm{~m}$
$\because$ Electric field on axis of dipole,
$E=\frac{4 k q a r}{\left(r^2-a^2\right)^2}$
where, $k$ is Coulomb's constant $=9 \times 10^9 \mathrm{C}^2 \mathrm{~m}^{-2} \mathrm{~N}^{-1}$
$\begin{aligned}
& 4 \times 9 \times 10^9 \times 500 \times 10^{-6} \\
& \Rightarrow \quad E=\frac{\times 5 \times 10^{-2} \times 25 \times 10^{-2}}{\left[\left(25 \times 10^{-2}\right)^2-\left(5 \times 10^{-2}\right)^2\right]^2} \\
& =\frac{2.25 \times 10^5}{[0.0625-0.0025]^2} \\
& =\frac{2.25 \times 10^5}{(0.06)^2} \\
& =\frac{2.25 \times 10^5}{3.6 \times 10^{-3}} \\
& =6.25 \times 10^7 \\
& \approx 6 \times 10^7 \mathrm{NC}^{-1} \\
&
\end{aligned}$
The result is close to option (a).