Case: Read the passage and answer the question.
When an excess charge is placed on an isolated conductor, it distributes itself on the surface of that conductor so that all points of the conductor, whether on the surface or inside the conductor, come to the same potential. This is true even if there is a cavity inside the conductor.
If an isolated conductor is placed in an external electric field, all points of the conductor still come to a single potential, regardless of whether the conductor has an excess charge. The free electrons distribute themselves on the surface in such a way that the electric field they produce at interior points cancels the external field that would otherwise be there.
An isolated metal sphere of radius 10 cm is charged to a potential of 900 V. What is the magnitude of the electric field at a point 20 cm from its centre?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2.25\times10^{3}\ \mathrm{V\,m^{-1}}\)
The charge on a metal sphere sits on its surface, so the sphere acts like a charged spherical shell. Outside it, the field and the potential are those of a point charge \( q \) at the centre. The surface potential \( V=q/4\pi\varepsilon_0R \) gives \( q/4\pi\varepsilon_0=VR \). The field at a distance \( r \) is then \( E=VR/r^{2} \).
Given:
\( V=900\ \mathrm{V} \)
\( R=10\ \mathrm{cm}=0.1\ \mathrm{m} \)
\( r=20\ \mathrm{cm}=0.2\ \mathrm{m} \)
\( \begin{aligned} E &= \frac{VR}{r^{2}} \\ &= \frac{900\times0.1}{(0.2)^{2}} \\ &= 2.25\times10^{3}\ \mathrm{V\,m^{-1}} \end{aligned} \)
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.”
Potential Due to a System of Charges – “The electric field inside the shell is zero. This implies … that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is”
\( \displaystyle V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R} \)\( \qquad [2.19(b)] \)