KCET · Physics · Electrostatics
Two metal spheres of radii \(0.01 \mathrm{~m}\) and \(0.02 \mathrm{~m}\) are given a charge of \(15 \mathrm{mC}\) and \(45 \mathrm{mC}\) respectively. They are then connected by a wire. The final charge on the first sphere is \(\ldots \ldots \ldots . \times 10^{-3} \mathrm{C}\).
- A 40
- B 30
- C 20
- D 10
Answer & Solution
Correct Answer
(C) 20
Step-by-step Solution
Detailed explanation
When two metal spheres are joined by a wire, charge flows from one at higher potential to the other at lower potential, till they acquire the same potential.
\(\begin{aligned} \quad V_{1} &=V_{2} \\ \frac{q_{1}}{4 \pi \varepsilon_{0} r_{1}} &=\frac{q_{2}}{4 \pi \varepsilon_{0} r_{2}} \end{aligned}\)
\(\Rightarrow \quad \frac{r_{1}}{r_{2}}=\frac{q_{1}}{q_{2}}=\frac{15 m C}{45 m C}=\frac{1}{3}\)
Final charge on first sphere,
\(q_{1}=\frac{1}{3}\left(q_{1}+q_{2}\right)=\frac{1}{3} \times 60=20 \mathrm{mC}\)
\(\begin{aligned} \quad V_{1} &=V_{2} \\ \frac{q_{1}}{4 \pi \varepsilon_{0} r_{1}} &=\frac{q_{2}}{4 \pi \varepsilon_{0} r_{2}} \end{aligned}\)
\(\Rightarrow \quad \frac{r_{1}}{r_{2}}=\frac{q_{1}}{q_{2}}=\frac{15 m C}{45 m C}=\frac{1}{3}\)
Final charge on first sphere,
\(q_{1}=\frac{1}{3}\left(q_{1}+q_{2}\right)=\frac{1}{3} \times 60=20 \mathrm{mC}\)
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