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GUJCET · Physics · Electric Charges and Fields
Three identical charges are placed on three vertices of a square. If the force acting between \(q_1\) and \(q_2\) is \(F _{12}\) and between \(q_1\) and \(q_3\) is \(F _{13}\) the \(\frac{F_{13}}{F_{12}}=\)_________.
- A \(\frac{1}{2}\)
- B 2
- C \(\frac{1}{\sqrt{2}}\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
(A)

Coloumb force acting between \(q_1\) and \(q_2\) \(
F_{12}=\frac{k q_1 q_2}{r^2}=\frac{k q^2}{r^2} \quad \quad \ldots \ldots(1)
\)
Coloumb force acting between \(q_1\) and \(q_3\) \(
\begin{array}{l}
F_{13}=\frac{k q_1 q_3}{r^2} \\
F_{13}=\frac{k q^2}{(\sqrt{2} r)^2} \quad(\because r=\sqrt{2} r)
\end{array}
\)
\(
F_{13}=\frac{k q^2}{2 r^2}\quad \quad \ldots \ldots(2)
\)
Taking the ratio of equation (2) and equation (1), \(
\frac{F_{13}}{F_{12}}=\frac{k q^2}{2 r^2} \times \frac{r^2}{k q^2}=\frac{1}{2}
\)

Coloumb force acting between \(q_1\) and \(q_2\) \(
F_{12}=\frac{k q_1 q_2}{r^2}=\frac{k q^2}{r^2} \quad \quad \ldots \ldots(1)
\)
Coloumb force acting between \(q_1\) and \(q_3\) \(
\begin{array}{l}
F_{13}=\frac{k q_1 q_3}{r^2} \\
F_{13}=\frac{k q^2}{(\sqrt{2} r)^2} \quad(\because r=\sqrt{2} r)
\end{array}
\)
\(
F_{13}=\frac{k q^2}{2 r^2}\quad \quad \ldots \ldots(2)
\)
Taking the ratio of equation (2) and equation (1), \(
\frac{F_{13}}{F_{12}}=\frac{k q^2}{2 r^2} \times \frac{r^2}{k q^2}=\frac{1}{2}
\)
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