Question

Two point charges $q$ and $-q$ are at positions $(0,0,d)$ and $(0,0,-d)$ respectively. What will be the electric field at point $(a,0,0)$ due to both the charges?

• A. $\frac{2kqd}{4\pi {\in }_0(d^2+a^2{)}^{3/2}}\hat{k}$
• B. $\frac{qd}{4\pi {\in }_0(d^2+a^2{)}^{3/2}}\hat{k}$
• C. $\frac{-2kqd}{4\pi {\in }_0(d^2+a^2{)}^{3/2}}\hat{k}$
• D. $\frac{-kqd}{4\pi {\in }_0(d^2+a^2{)}^{3/2}}\hat{k}$

Answer 1

The correct option is C $\frac{-2kqd}{4\pi {\in }_0(d^2+a^2{)}^{3/2}}\hat{k}$

At P there are two fields,
$\left(1\right)$ Due to $A,{\overrightarrow{E}}_A=\frac{kq}{r^2}$ Away from A
$\left(2\right)$ Due to $B,{\overrightarrow{E}}_B=\frac{kq}{r^2}$ Towards B
Distance of point P from both A and B, $r=(a^2+d^2{)}^{1/2}$
Angle made by ${\overrightarrow{E}}_{net}$ with ${\overrightarrow{E}}_A$ or ${\overrightarrow{E}}_B$ $=\frac{1}{2}(180-2\theta )=(90-\theta )$
Net electric field at P, ${\overrightarrow{E}}_{net}=2E_A\cos (90-\theta )$
$\Rightarrow {\overrightarrow{E}}_{net}=2E_A\sin \theta =\frac{2kq}{r^2}\frac{d}{\left(r\right)}$
${\overrightarrow{E}}_{net}=\frac{2kqd}{(a^2+d^2{)}^{3/2}}$ in $-z$ direction
${\overrightarrow{E}}_{net}=\frac{-2kqd}{(a^2+d^2{)}^{3/2}}\hat{k}$