Question

Two unit negative charges are placed on a straight line. A positive charge $q$ is placed exactly at the midpoint between these unit charges. If the system of these three charges is in equilibrium, the value of $q\left(inC\right)$ is :

• A. $1.0$
• B. $0.75$
• C. $0.5$
• D. $0.25$

Answer 1

The correct option is D $0.25$

Since the positive charge is equidistant from the equal negative charges, the net force on it is zero. Let the distance between the negative charges be ${}^{\prime }{r}^{\prime }$, as shown in the figure. For equilibrium, the net force on each $-1C$ charge is zero.
$⟹$(force on $-1C$ due to other $-1C)+($force on $-1C$ due to $q)=0$
$\Rightarrow \frac{1}{4\pi \epsilon o}\frac{(-1C)\times (-1C)}{r^2}+\frac{1}{4\pi \epsilon o}\frac{(-1C)\times q}{(\frac{r}{2}{)}^2}=0$
$\Rightarrow \frac{1}{r^2}+\frac{-q}{\frac{r^2}{4}}=0$
$\Rightarrow 1-4q=0$
$q=0.25C$