Question

The volume charge density of a solid sphere of radius 0.75 m is 0.25 nCm^−3. Apply Gauss’s law to calculate the a) electric flux through the sphere and b) electric field at a distance of 1.5 m and at a distance of 0.50 m from the centre of the sphere, respectively. If this sphere were made of conducting material, what would the values of the electric fields be at these distances?

Answer 1

(a) The electric flux

$$\phi=\oiint \bf{E}\cdot d\bf{A}$$

The Gauss’s law states

$$\oiint {\bf E}\cdot d{\bf A}=\iiint {\rm div} {\bf E}\: dV=\frac{1}{\varepsilon_0}\iiint \rho\: dV$$

So, the electric flux

$$\phi=\frac{1}{\varepsilon_0}\iiint \rho\: dV=\frac{\rho}{\varepsilon_0}\times \frac{4}{3}\pi R^3$$

$$\phi=\frac{0.25\times 10^{-9}}{8.85\times 10^{-12}}\times \frac{4}{3}\pi (0.75)^3=50\:\rm{V\cdot m}$$

(b) If $r>R$ the electric field from both dielectric and conductive sphere

$$\oiint {\bf E}\cdot d{\bf A}=\frac{1}{\varepsilon_0}\iiint \rho\: dV$$

$$E\times 4\pi r^2=\frac{\rho}{\varepsilon_0}\times \frac{4}{3}\pi R^3$$

$$E=\frac{\rho R^3}{3\varepsilon_0 r^2}=\frac{0.25\times 10^{-9}\times (0.75)^3}{3\times 8.85\times 10^{-12}\times (1.5)^2}=1.77\:\rm{V/m}$$

If $r<R$ the electric field from dielectric sphere

$$E\times 4\pi r^2=\frac{\rho}{\varepsilon_0}\times \frac{4}{3}\pi r^3$$

$$E=\frac{\rho r}{3\varepsilon_0}=\frac{0.25\times 10^{-9}\times 0.50}{3\times 8.85\times 10^{-12}}=4.71\:\rm{V/m}$$

If $r<R$ the electric field inside a conductive sphere is zero. This is because all charge is at the surface of sphere.