Question

Two point charges $+8q$ and $-2q$ are located at x $=$ 0 and x $=$ L respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero :

• A. $2L$
• B. $\frac{L}{4}$
• C. $8L$
• D. $4L$

Answer 1

Answer: (A)

$2L$

Consider the region between $x=-\infty$ and $x=0$.

The electric field due to $+8q$ is towards left and that due to $-2q$ is towards right.

Electric field at a general point $x=d(<0)$ is $E=\frac{1}{4\pi {ϵ}_0}(-\frac{8q}{d^2}+\frac{2q}{(L-d{)}^2})$

If $E=0$, then $\frac{8q}{d^2}=\frac{2q}{(L-d{)}^2}\Rightarrow \frac{d^2}{(L-d{)}^2}=4$

But, if $d<0$, then $|d|<|L-d|$. Hence, no solution exists.

Consider the region between $x=0$ and $x=L$

The electric field due to $+8q$ is towards right, and that due to $-2q$ is towards right.

Hence, there is no point in$(0,L)$ with zero electric field.

Consider the region between $x=L$ and $x=+\infty$

The electric field due to $+8q$ is towards right and $-2q$ is towards left.

Electric field at a general point $x=d(>L)$ is given by $E=\frac{1}{4\pi {ϵ}_0}(+\frac{8q}{d^2}-\frac{2q}{(d-L{)}^2})$

If $E=0$, then $\frac{8q}{d^2}=\frac{2q}{(d-L{)}^2}\Rightarrow \frac{d}{d-L}=\pm 2\Rightarrow d=2L\text{or}d=2L/3$

But, $d>L$.

Hence $d=2L$

Hence at $x=2L$, net electric field is zero.