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 ,
$\vec{E}=\overrightarrow{E_1}+\overrightarrow{E_2}=0$
$|\vec{E}|=\frac{1}{4\pi {\in }_0}\frac{\frac{q_1}{2}}{r_1}+\frac{1}{4\pi {\in }_0}\frac{\frac{q_2}{2}}{r_2^2}$
$\Rightarrow 0=\frac{1}{4\pi {\in }_0}[\frac{4\mu C}{x^2}+\frac{(-2\mu C)}{(1-x^2)}]$
$\Rightarrow \frac{4\mu C}{x^2}=\frac{2\mu C}{(1-x^2)}\Rightarrow \frac{2}{x^2}=\frac{1}{(1-x^2)}$
$\Rightarrow 2(1-x{)}^2=x^2$
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