Question

The radius of a hypothetical nucleus (atomic number $=79$) is about $7\times {10}^{-15}$ m. Assuming that charge distribution is uniform, the electric field at distance of $3.5\times {10}^{-15}$ of the nucleus is :

• A. $3\times {10}^{20}$
• B. $1.7\times {10}^{21}$
• C. $3.5\times {10}^{20}$
• D. $3.5\times {10}^{21}$

Answer 1

The correct option is B $1.7\times {10}^{21}$

Observe the figure. The nucleus is a uniformly distributed sphere of charge of radius $a=7\times {10}^{-15}\text{m}$

Consider a Gaussian sphere as shown in the figure, centred at the centre of nucleus, with radius of $r=3.5\times {10}^{-15}\text{m}$

By the radial symmetry of the system, electric field field is in radial direction i.e., $\overrightarrow{E}=E\left(r\right)\hat{r}$

The total flux through the Gaussian sphere is $\oint \overrightarrow{E}.\overrightarrow{dS}=\oint E\left(r\right)\hat{r}.dS\hat{r}=4\pi r^{2}E\left(r\right)$

But, by Gauss' theorem, flux is $\frac{q_{\text{encl}}}{{ϵ}_0}=\frac{4}{3}\pi r^3\frac{\rho }{{ϵ}_0}=\frac{4\pi r^3}{3{ϵ}_0}\frac{Q_{\text{total}}}{V_{\text{total}}}=\frac{Q_{\text{total}}}{{ϵ}_0}\frac{r^3}{a^3}$

$Q_{\text{total}}=79|e|=79\times 1.602\times {10}^{-19}\text{C}=1.266\times {10}^{-17}\text{C}$

Thus, $4\pi r^{2}E=\frac{r^3{Q}_{\text{total}}}{{ϵ}_0{a}^3}\Rightarrow E=\frac{Q_{\text{total}}r}{4\pi {ϵ}_0{a}^3}=\frac{1.266\times {10}^{-17}\times 3.5\times {10}^{-15}\times 9\times {10}^9}{(7\times {10}^{-15}{)}^3}\approx 1.7\times {10}^{21}\text{N/C}$