Question

Two point charges 4$\mu$C and - 2$\mu$C are separated by a distance of 1 m in air. Then the distance of the point on the line joining the charges, where the resultant electric field is zero, is (in metre)

• A. 0.58
• B. 0.75
• C. 0.67
• D. 0.81

Answer 1

Answer: (A)

Let the point P where resultant field is zero be x m from 4$\mu$C charge and (1- x) m distance apart from -2$\mu$C charge. Since field is zero at this point then ,
$\vec{E} \, =\vec{E_1}+\vec{E_2} \, = \, 0$
$|\vec{E}| \, = \, \frac{1}{4\pi\in_0}\frac{\frac{q_1}{2}}{r_1}+\frac{1}{4\pi\in_0}\frac{\frac{q_2}{2}}{r^2_2}$
$\Rightarrow 0 =\frac{1}{4\pi \in_0 }\bigg[ \frac{4\mu C}{x^2}+\frac{(-2\mu C)}{(1-x^2)} \bigg]$
$\Rightarrow \frac{4\mu C}{x^2}=\frac{2\mu C}{(1-x^2)} \Rightarrow \frac{2}{x^2}=\frac{1}{(1-x^2)}$
$\Rightarrow 2(1-x)^2 \, = \, x^2$
Taking Root $\sqrt{2}(1-x) \, = \, x$
$\Rightarrow $ 1.414(1-x) \, = \, x $\Rightarrow $ 1.414-1.414x \, = \, x
$\Rightarrow $ 1.414 \, = \, (1+414)x $\Rightarrow $x= $\frac{1.414}{2.414}$
$\Rightarrow $ x \, = \, 0.58m