Q150 / 160 · AP EAMCET 2007 · Physics
Two unit negative charges are placed on a straight line. A positive charge $q$ is placed exactly at the mid point between these unit charges. If the system of these three charges is in equilibrium, the value of $q$ (in $\mathrm{C}$ ) is
- A$1.0$
- B$0.75$
- C$0.5$
- Correct answerD$20 \mathrm{~cm}$
Solution
For equilibrium, we have
$F_{A B}+F_{A C}=0$ or $\quad \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q}{(d / 2)^2}+\frac{1}{4 \pi \varepsilon_0} \times \frac{q_1 q_2}{d^2}=0$ Given, $\quad q_1=q_2=-1 \mu \mathrm{C}$ So, $\quad-\frac{q}{(d / 2)^2}+\frac{1}{d^2}=0$ $q=\frac{1}{4}=0.25 \mathrm{C}$
$F_{A B}+F_{A C}=0$ or $\quad \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q}{(d / 2)^2}+\frac{1}{4 \pi \varepsilon_0} \times \frac{q_1 q_2}{d^2}=0$ Given, $\quad q_1=q_2=-1 \mu \mathrm{C}$ So, $\quad-\frac{q}{(d / 2)^2}+\frac{1}{d^2}=0$ $q=\frac{1}{4}=0.25 \mathrm{C}$
Hence, the correct answer is option: 'D'.