Question

The radius of a gold nucleus $(z=79)$ is about $7.0\times {10}^{-15}m.$. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) The surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?

Answer 1

Charge present in a gold nucleus $=79\times 1.6\times {10}^{-19}C$

Since the surface encloses all the charges we have:

(a) $\oint \overrightarrow{E}\overrightarrow{ds}=\frac{q}{{\in }_0}=\frac{79\times 1.6\times {10}^{-19}}{8.85\times {10}^{-12}}$

$E=\frac{q}{{\in }_{0}ds}=\frac{79\times 1.6\times {10}^{-19}}{8.85\times {10}^{-12}}\times \frac{1}{4\times 3.14\times (7\times {10}^{-15}{)}^2}$ $[\because area=4\pi r^2]$

$=2.32\times {10}^{21}N/C$

(b) For the middle part of the radius, Now here $r=7/2\times {10}^{-15}m$

Volume $=4/3\pi r^3=\frac{48}{3}\times \frac{22}{7}\times \frac{343}{8}\times {10}^{-45}$

Charge enclosed $=\zeta \times volume$ [$\zeta :$ volume chage density]

But $\zeta =\frac{\text{Net charge}}{\text{Net volume}}=\frac{7.9\times 1.6\times {10}^{-19}C}{\left(\frac{4}{3}\right)\times \pi \times 343\times {10}^{-45}}$

Net charged enclosed $=\frac{\text{Net charge}}{\text{Net volume}}$

$=\frac{7.9\times 1.6\times {10}^{-19}C}{\left(\frac{4}{3}\right)\times \pi \times 343\times {10}^{-45}}\times \frac{4}{3}\pi \times \frac{343}{8}\times {10}^{-45}=\frac{7.9\times 1.6\times {10}^{-19}}{8}$

$\oint \overrightarrow{E}\overrightarrow{ds}=\frac{q_{enclosed}}{{\in }_0}$

$=E=\frac{7.9\times 1.6\times {10}^{-19}}{8\times {\in }_0\times S}=\frac{7.9\times 1.6\times {10}^{-19}}{8\times 8.85\times {10}^{-12}c\times 4\pi \times \frac{49}{4}\times {10}^{-30}}=1.159\times {10}^{21}N/C$