TS EAMCET · Physics · Electrostatics
A thin metallic spherical shell of radius \(r\) contains a charge \(Q\) on its surface. A point charge \(q_1\) is placed at the centre of shell and another charge \(q_2\) is placed outside the shell at a distance \(x\) from the centre. Then, the forces on charges \(q_1\) and \(q_2\) respectively are
- A \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}, \frac{1}{4 \pi \varepsilon_0} \frac{q_1+q_2}{x^2}\)
- B \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}, \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{x^2}\)
- C \(0, \frac{q_2}{4 \pi \varepsilon_0} \frac{Q}{x^2}\)
- D \(0, \frac{q_2}{4 \pi \varepsilon_0} \frac{Q+q_1}{x^2}\)
Answer & Solution
Correct Answer
(D) \(0, \frac{q_2}{4 \pi \varepsilon_0} \frac{Q+q_1}{x^2}\)
Step-by-step Solution
Detailed explanation
Given situation is As field inside shell is zero, so force on \(q_1=q_1 E_{\mathrm{in}}=0\). Also force on \(q_2\) is due to both \(q_1\) and \(Q\). \(F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2 Q}{x^2}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2 q_1}{x^2}\)…
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