A thin spherical conducting shell of radius $\mathrm{R}$ has a charge $\mathrm{q}$. Another charge…

A thin spherical conducting shell of radius $\mathrm{R}$ has a charge $\mathrm{q}$. Another charge $\mathrm{Q}$ is placed at the centre of the shell. The electrostatic potential at a point $\mathrm{P}$ a distance $\frac{\mathrm{R}}{2}$ from the centre of the shell is
  1. $\frac{2 Q}{4 \pi \varepsilon_0 R}$
  2. $\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}}-\frac{2 \mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}$
  3. $\frac{2 Q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 R}$
  4. $\frac{(\mathrm{q}+\mathrm{Q}) 2}{4 \pi \varepsilon_0 \mathrm{R}}$

Solution

Potential due to spherical shell, $\mathrm{v}_1=\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}$. Potential difference due to charge at the centre $\mathrm{V}_2=\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 r} ; \mathrm{V}=\mathrm{V}_1+\mathrm{V}_2=\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}}+\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}$

Asked in: JEE Main 2003

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