91Ó°ÊÓ

A uniformly charged insulating sphere that has volume charge density \(\rho \) and radius \(R\)

It states that the line integral of the electric field over a closed surface is equal to \(\frac{1}{{{\varepsilon _0}}}\) times the total charge enclosed by the closed surface. It is expressed as,

\(\int {\vec E \cdot d\vec s} = \frac{{{q_{{\rm{enc}}}}}}{{{\varepsilon _0}}}\) …(i)

Where, \({\varepsilon _0}\) is permittivity of free space, \({q_{{\rm{enc}}}}\) is the charge enclose by the Gaussian surface, \(\vec E\) is the electric field vector and \(d\vec s\) is the surface area vector

Substitute \(d\vec s = 4\pi {R^2}\)and \({q_{{\rm{enc}}}} = \rho \frac{4}{3}\pi {R^3}\)in equation (i)

\(\begin{aligned}E(4\pi {R^2}) = \frac{{\rho \frac{4}{3}\pi {R^3}}}{{{\varepsilon _0}}}\\E = \frac{\rho }{{3{\varepsilon _0}}}r\end{aligned}\)

Then the slope between \(E\) and \(r\) will be

\({\rm{slope}} = \frac{\rho }{{3{\varepsilon _0}}}\) …(ii)

From the graph, slope of the graph is calculated as,

\(\begin{aligned}{\rm{slope}}\, &= \,\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\\ &= \frac{{6.0 \times {{10}^4}{\rm{N/C}} - 3.0 \times {{10}^4}{\rm{N/C}}}}{{8.0 \times {{10}^{ - 3}}\,{\rm{m}} - 4.0 \times {{10}^{ - 3}}\,{\rm{m}}}}\\ &= 7.5 \times {10^6}{\rm{N/mC}}\end{aligned}\) …(iii)

By equation (ii) and (iii)

\(\begin{aligned}\frac{\rho }{{3{\varepsilon _0}}} &= 7.5 \times {10^6}{\rm{N/mC}}\\\rho &= 3{\varepsilon _0} \times 7.5 \times {10^6}{\rm{N/mC}}\end{aligned}\)

Substitute \({\varepsilon _0} = 8.85 \times {10^{ - 12}}{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}\)in above equation

\(\begin{aligned}\rho &= 3\left( {8.85 \times {{10}^{ - 12}}{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}} \right) \times \left( {7.5 \times {{10}^6}{\rm{N/mC}}} \right)\\ &= 1.99 \times {10^{ - 4}}{\rm{C/}}{{\rm{m}}^{\rm{3}}}\end{aligned}\)

Hence, the volume charge density is\(\rho = 1.99 \times {10^{ - 4}}{\rm{C/}}{{\rm{m}}^{\rm{3}}}\)