TS EAMCET · Physics · Electrostatics
A conducting sphere \(S_1\) of radius \(r_1\) is connected by a conducting wire to another conducting sphere \(S_2\) of radius \(r_2\), where \(r_1=3 \mathrm{~cm}\) and \(r_2=2\) \(\mathrm{cm}\). Before they are connected, \(s_1\) carries charge of 10 units. The electric potential at the point which is at a distance \(4 \mathrm{~cm}\) from the centre of \(S_1\) and a distance \(3 \mathrm{~cm}\) from the centre of \(S_2\) is
- A \(\frac{1}{4 \pi \varepsilon_0} \frac{17}{6}\)
- B \(\frac{1}{4 \pi \varepsilon_0} \frac{3}{2}\)
- C \(\frac{1}{4 \pi \varepsilon_0} \frac{1}{6}\)
- D \(\frac{1}{4 \pi \varepsilon_0} \frac{17}{12}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4 \pi \varepsilon_0} \frac{17}{6}\)
Step-by-step Solution
Detailed explanation
When spheres are connected, charges are redistributed such that potential on their surfaces are same. So, we have \( \begin{aligned} q_1+q_2 & =10 \\ \frac{k q_1}{r_1} & =\frac{k q_2}{r_2} \Rightarrow \frac{q_1}{3}=\frac{q_2}{2} \end{aligned} \) and…
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