A thin spherical shell of radius 9 cm carries charge spread uniformly over it, with a surface charge density of \(8.85\times10^{-9}\ \mathrm{C\,m^{-2}}\). The electric potential at the surface of the shell is (Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(90\ \mathrm{V}\)
Outside a uniformly charged shell, and on its surface, the potential is as if the whole charge were at the centre. So at the surface \( V=q/4\pi\varepsilon_0R \). The total charge is the surface charge density times the area \( 4\pi R^{2} \).
\( \begin{aligned} V &= \frac{q}{4\pi\varepsilon_0R} \\ &= \frac{\sigma\times4\pi R^{2}}{4\pi\varepsilon_0R} \\ &= \frac{\sigma R}{\varepsilon_0} \end{aligned} \)
\( \begin{aligned} V &= \frac{8.85\times10^{-9}\times0.09}{8.85\times10^{-12}} \\ &= 90\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Potential Due to a System of Charges – “We have seen in Chapter 1 that for a uniformly charged spherical shell, the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by”
\( V=\frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}\quad(r\ge R) \)\( \qquad (2.19(a)) \)
“where q is the total charge on the shell and R its radius. …”