Question

A Hollow charged metal sphere has radius $r$ . If the potential difference between its surface and a point at a distance $3r$ from the center is $V$ , then the electric field intensity at a distance $3r$ from the center is.

Answer 1

The hollow charged sphere is electrically charged and placed at a point. Therefore, we can apply the law of electrostatics that is Coulomb’s law here in this question. By using Coulomb’s law we can find the electric field at the surface of the sphere and at the point where we have to find.

Complete step by step answer:
Given charge is distributed on the hollow sphere uniformly. Let that charge be ${\text{Q}}$
This means that the potential and electric field outside the sphere are the same as these quantities for a point charge ${\text{Q}}$ placed at the center of the sphere.
Thus, the potential difference between the surface of the sphere and the point at a distance of $3r$ is
$V{\text{ = }}\dfrac{1}{{4\pi {\varepsilon _0}}}(\dfrac{{\text{Q}}}{r} - \dfrac{{\text{Q}}}{{3r}}){\text{ = }}\dfrac{2}{3}\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\text{Q}}}{r}{\text{ }}..........{\text{(i)}}$
The electric field at the point of distance $3r$:
$E{\text{ = }}\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\text{Q}}}{{{{(3r)}^2}}}\, = \,\dfrac{1}{9}\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\text{Q}}}{{{r^2}}}\,\, = \,\dfrac{1}{{9r}}\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\text{Q}}}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,........({\text{ii}})$
From $({\text{i}})$ and $({\text{ii}})$
$E{\text{ = }}\dfrac{1}{{9r}}\dfrac{{3V}}{2}\, = \dfrac{V}{{6r}}$
Hence, $E{\text{ = }}\dfrac{V}{{6r}}$ is the answer.

Note: Due to absence of the charge inside the hollow sphere the electric field inside the hollow sphere is always zero. Because charges inside the gaussian surface formed will be zero. When we talk outside the sphere there will be the whole sphere in the gaussian surface that’s why we can take it as a point charge to calculate the electric field and the potential difference outside the sphere.