Question

The magnitude of the electric field as a function of the distanee r outside the ball is given by :

• A. $E=\frac{{\rho }_0{R}^3}{8\epsilon r^2}$
• B. $E=\frac{{\rho }_0{R}^3}{12\epsilon r^2}$
• C. $E=\frac{{\rho }_0{R}^2}{8\epsilon r^2}$
• D. $E=\frac{{\rho }_0{R}^2}{12\epsilon r^2}$

Answer 1

The correct option is B $E=\frac{{\rho }_0{R}^3}{12\epsilon r^2}$

The charge of the sphere is $q={\int }_0^R\rho \left(4\pi r^2\right)dr=4\pi {\rho }_0{\int }_0^R[1-r/R]r^{2}dr=4\pi {\rho }_0[r^3/3-r^4/4R{]}_0^R=4\pi {\rho }_0[R^3/3-R^4/4R]=4\pi {\rho }_0(R^3/12)$
Using Gauss's law the field at a distance r outside the sphere is
$E.4\pi r^2=\frac{Q_{en}}{ϵ}=\frac{q}{ϵ}=\frac{4\pi {\rho }_0\left(R^3/12\right)}{ϵ}$
or $E=\frac{{\rho }_0{R}^3}{12ϵr^2}$