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 gold nucleus

$=739\times 1.6\times 10-19C$

Since the surface enclose all the charges, we have

(a) $E\times ds=\frac{Q}{{\in }_0}$

$E=\frac{Q}{{\in }_{0}ds}=x[area=4\pi r^2]$

$=2.315131\times {10}^{21}N/C$

(b) For the middle point of the radius,

NOw, here $r=\frac{7}{2}\times {10}^{15}m$

volume $=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times \frac{22}{7}\times \frac{348}{8}\times {10}^{-45}$

Net charge $=79\times 1.6\times {10}^{-19}C$

Volume charge density

$=\frac{79\times 1.6\times {10}^{-19}}{\frac{4}{3}\pi \times 343\times {10}^{-45}}$

So Charge m required part

$=\frac{79\times 1.6\times {10}^{-19}}{\frac{4}{3}\pi \times 343\times {10}^{-45}}\times \frac{4}{3}\pi \times \frac{343}{8}\times {10}^{-45}$

$\frac{79\times 1.6\times {10}^{-19}}{8}$

So, $\frac{79\times 1.6\times {10}^{-19}}{4\pi {\in }_0.r^2}$

$1.16\times {10}^{21}N/C.$