TS EAMCET · Physics · Electrostatics
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 \(0.5\)
- D \(20 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(20 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
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, \(\quad-\frac{q}{(d / 2)^2}+\frac{1}{d^2}=0\)…
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