Question

Two equal negative point charges $-q$ are fixed at points $(0,a)$ and $(0,-a)$ on the $\text{y-}$axis. A positive point charge $Q$ is in rest at point $(2a,0)$ on the $\text{x-}$axis. The net electric field on $Q$ due to both $-q$ charges will be $(\frac{1}{4\pi {ϵ}_0}=k)$

• A. $\frac{4qk}{3a^2}$
• B. $\frac{qk}{a^2}$
• C. $\frac{4qk}{5\sqrt{5}{a}^2}$
• D. $\frac{4qk}{25a^2}$

Answer 1

The correct option is C $\frac{4qk}{5\sqrt{5}{a}^2}$

Let the charge $Q$ be at $\text{P}$, with $\text{OP}=x$. The resultant field $E_r$ is along the $\text{x-}$axis directed towards the origin.



Let, $\text{AP=BP}=r$.

Then, the electric fields due to both the cahrges,
$E_1=E_2=\frac{q}{4\pi {\epsilon }_0{r}^2}$

From figure, the resultant field on $Q$ is
$E_r=E_1\cos \theta +E_2\cos \theta$

Substituting the value of $E_1$ and $E_2$, we get
$E_r=\frac{q}{4\pi {\epsilon }_0{r}^2}\cos \theta +\frac{q}{4\pi {\epsilon }_0{r}^2}\cos \theta$

$\Rightarrow E_r=\frac{2q}{4\pi {\epsilon }_0{r}^2}\cos \theta$

Substituting the value of $\cos \theta$ from figure,
$\Rightarrow E_r=\frac{2q}{4\pi {\epsilon }_0}\frac{x}{{(a^2+x^2)}^{\frac{3}{2}}}$

As, $x=2a$

$\Rightarrow E_r=\frac{2q}{4\pi {\epsilon }_0}\frac{2a}{{(a^2+(2a{)}^2)}^{\frac{3}{2}}}$

$\Rightarrow E_r=\frac{4qak}{(5a^2{)}^{\frac{3}{2}}}(\because \frac{1}{4\pi {ϵ}_0}=k)$

$\therefore E_r=\frac{4qk}{5\sqrt{5}{a}^2}$

Hence, option (c) is the correct answer.