MHT CET · Physics · Electrostatics
Three equal charges \(q_1, q_2\) and \(q_3\) are placed on the three corners of a square of side \(a\). If the force between \(q_1\) and \(q_2\) is \(F_{12}\) and that between \(q_1\) and \(q_3\) is \(F_{13}\), then the ratio of magnitudes \(\left(\frac{F_{12}}{F_{13}}\right)\) is
- A \(\frac{1}{\sqrt{2}}\)
- B \(\frac{1}{2}\)
- C 2
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
Suppose the side of square is \(a\). Then the diagonal is equal to \(\sqrt{2} a\)
\(\begin{aligned} & \therefore F_{12}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{a^2} \\ & \text { and } F_{13}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_3}{(\sqrt{2} a)^2}\end{aligned}\)
We know, \(q_1=q_2=q_3=q\)
So, \(\frac{F_{12}}{F_{13}}=\frac{2 q_2}{q_3}=2\)
\(\begin{aligned} & \therefore F_{12}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{a^2} \\ & \text { and } F_{13}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_3}{(\sqrt{2} a)^2}\end{aligned}\)
We know, \(q_1=q_2=q_3=q\)
So, \(\frac{F_{12}}{F_{13}}=\frac{2 q_2}{q_3}=2\)
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