Question

The radius of gold nucleus is about $7\times {10}^{-15}m(Z=79).$ The electric field at the mid-point of the radius assuming charge is uniformly distributed is :

• A. $1.16\times {10}^{19}N{C}^{-1}$
• B. $1.16\times {10}^{21}N{C}^{-1}$
• C. $2.32\times {10}^{21}N{C}^{-1}$
• D. $2.32\times {10}^{19}N{C}^{-1}$

Answer 1

The correct option is B $1.16\times {10}^{21}N{C}^{-1}$

Let $r$ be the radius of the gold and $E$ be the electric field at $r/2$.

If $\rho$ is the volume charge density, $Q=Ze=\rho .\frac{4}{3}\pi r^3...\left(1\right)$

using Gauss's law, $E.4\pi (r/2{)}^2=\frac{Q_{enclosed}}{{ϵ}_0}=\frac{\rho .\frac{4}{3}\pi (r/2{)}^3}{{ϵ}_0}$

or $E=\frac{Ze}{8\pi {ϵ}_0{r}^2}$ using $\left(1\right)$

$E=\frac{79\times 1.6\times {10}^{-19}}{8\times 3.14\times 8.854\times {10}^{-12}\times 49\times {10}^{-30}}=1.16\times {10}^{21}N{C}^{-1}$