Question

$N$ negative point charges, each $-q/N$, are kept of radius ${}^{\prime }{a}^{\prime }$. $P$ is a point on a line perpendicular to the plane of circle and passing through its centre, distance of $P$from the centre being ${}^{\prime }{Y}^{\prime }$. Magnitude of electric field strength at $P$ is

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{q}{x^2+a^2}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{nq}{(y^2+a^2{)}^{3/2}}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{qy}{(y^2+a^2{)}^{3/2}}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{nqy}{(y^2+a^2{)}^{3/2}}$

Answer 1

The correct option is A $\frac{1}{4\pi {\epsilon }_0}\frac{q}{x^2+a^2}$

To determine the electric field which result from the dipole charges in fig that are separated by a distance $2G$, We simply sum the fields from each charge individually. The point to noted that the fields are vectors.

$\overline{E}={\overline{E}}_1+{\overline{E}}_2$

The distance of each charge $\left(r\right)$ from the point where the electric field is measured is $r=\sqrt{a^2+x^2}$.

The magnitude of the electric field $\left(E_1\right)$ due to the first charge $\left(q_1\right)$ and the magnitude of the electric field $\left(E_2\right)$ due to the $2nd$ charge $\left(q_2\right)$ are the same although the direction are different.

$E_1=E_2=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

$=\frac{1}{4\pi {ϵ}_0}\frac{q}{a^2+x^2}$