MHT CET · Physics · Electrostatics
A solid metallic sphere has a charge \(+3 Q\). Concentric with this sphere is a conducting spherical shell having charge \(-\mathrm{Q}\). The radius of the sphere is ' \(A\) ' and that of the spherical shell is ' \(\mathrm{B}\) '. \((\mathrm{B}>\mathrm{A})\). The electric field at a distance ' \(R\) ' ( \(A < R < B)\) from the centre is ( \(\varepsilon_0=\) permittivity of vacuum)
- A \(\frac{\mathrm{Q}}{2 \pi \varepsilon_0 \mathrm{R}}\)
- B \(\frac{\mathrm{3Q}}{2 \pi \varepsilon_0 \mathrm{R}}\)
- C \(\frac{\mathrm{3Q}}{4 \pi \varepsilon_0 \mathrm{R^2}}\)
- D \(\frac{\mathrm{4Q}}{2 \pi \varepsilon_0 \mathrm{R^2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{3Q}}{4 \pi \varepsilon_0 \mathrm{R^2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \therefore \quad & \text { Using } \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}, \text { we get } \\ & \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \frac{(3 \mathrm{Q})}{\mathrm{R}^2}\end{aligned}\)
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