Question

The maximum electric field intensity is :

• A. $E_m=\frac{{\rho }_{0}R}{9\epsilon }$
• B. $E_m=\frac{{\rho }_{0}R}{12\epsilon }$
• C. $E_m=\frac{{\rho }_{0}R}{3\epsilon }$
• D. $E_m=\frac{{\rho }_{0}R}{6\epsilon }$

Answer 1

Answer: (A)

$E_m=\frac{{\rho }_{0}R}{9\epsilon }$

Using Gauss's law the field at a distance r from center of the sphere is
$E.4\pi r^2=\frac{Q_{en}}{ϵ}$
here, $Q_{en}={\int }_0^r\rho \left(4\pi r^2\right)dr=4\pi {\rho }_0{\int }_0^r[1-r/R]r^{2}dr=4\pi {\rho }_0[r^3/3-r^4/4R{]}_0^r=4\pi {\rho }_0[r^3/3-r^4/4R]$
Thus, $E.4\pi r^2=\frac{4\pi {\rho }_0[r^3/3-r^4/4R]}{ϵ}$
or $E=\frac{{\rho }_0}{ϵ}[r/3-r^2/4R]...\left(1\right)$
For max E, $\frac{dE}{dr}=0=\frac{{\rho }_0}{ϵ}[1/3-2r/4R]$
or $1/3=r/2R\Rightarrow r=r_m=\frac{2R}{3}$
Putting the value $r=r_m=2R/3$ in (1), we get
$E_{max}=\frac{{\rho }_0}{ϵ}[\frac{2R/3}{3}-\frac{(2R/3{)}^2}{4R}]=\frac{{\rho }_{0}R}{ϵ}[2/9-1/9]=\frac{{\rho }_{0}R}{9ϵ}$