Question

Two point charges $ +8q$ and $-2q$ are placed as shown in the figure.

The point where the electric field is zero is

• A. $x = l$
• B. $x = 2l$
• C. $x = 3l$
• D. $x = 4l$

Answer 1

Answer: (B)

Let at a point $P$ at a distance $d$ from $-2q$ charge at which the electric field is zero.

$\therefore E_1 = E_2$
$\frac{1}{4\pi\varepsilon_{0}} \frac{8q}{\left(l+d\right)^{2}} = \frac{1}{4\pi\varepsilon_{0}} \frac{2q}{d^{2}}$
$ 4d^{2} = \left(l+d\right)^{2} $
$ 2d = l+d $
$d= l$
Hence, at $x = 2l$, the electric field is zero due to given charge configuration.