Three charges $+q,+2 q$ and $+4 q$ are connected by strings as shown in the figure. What is ratio of…
- $1: 2$
- $1: 3$
- $2: 1$
- $3: 1$
Solution
\(\mathrm{k} \frac{2 \mathrm{q}^2}{\mathrm{~d}^2}+\frac{4 \mathrm{q}^2}{4 \mathrm{~d}^2}=\mathrm{T}_{\mathrm{AB}} \Rightarrow \mathrm{T}_{\mathrm{AB}}=\frac{3 \mathrm{kq}^2}{\mathrm{~d}^2}\)From equilibrium of charge \(4 \mathrm{q}, \mathrm{k} \frac{8 \mathrm{q}^2}{\mathrm{~d}^2}+\frac{4 \mathrm{q}^2}{4 \mathrm{~d}^2}=\mathrm{T}_{\mathrm{BC}} \Rightarrow \mathrm{T}_{\mathrm{BC}}=\frac{9 \mathrm{kq}^2}{\mathrm{~d}^2}\)
Thus, \(\frac{\mathrm{T}_{A B}}{\mathrm{~T}_{B C}}=\frac{1}{3}\)
Asked in: JEE Mains - Electrostatics - Test 1