Question

The radius of the gold nucleus is about $7 \times 10^{-15} \mathrm{m}(Z=79)$. The electric field at the
mid-point of the radius assuming charge is uniformly distributed is :
A $1.16\times {{10}^{19}}\text{N}{{\text{C}}^{-1}}$
B $1.16 \times 10^{21} \mathrm{NC}^{-1}$
C $2.32 \times 10^{21} \mathrm{NC}^{-1}$
D $2.32 \times 10^{19} \mathrm{NC}^{-1}$

Answer 1

Hint The space around an electric charge in which its influence can be felt is known as the electric field. The electric field Intensity at a point is the force experienced by a unit positive charge placed at that point. Electric Field Intensity is a vector quantity. Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

Complete step by step answer
Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.

Let $\mathrm{r}$ be the radius of the gold and $\mathrm{E}$ be the electric field at $\mathrm{r} / 2$. If $\rho$ is the volume charge density, $\mathrm{Q}=\mathrm{Ze}=\rho \cdot \dfrac{4}{3} \pi \mathrm{r}^{3} \ldots(1)$
using Gauss's law, $\mathrm{E} .4 \pi(\mathrm{r} / 2)^{2}=\dfrac{\mathrm{Q}_{\text {enclosed }}}{\epsilon_{0}}=\dfrac{\rho \cdot \dfrac{4}{3} \pi(\mathrm{r} / 2)^{3}}{\epsilon_{0}}$
or $\mathrm{E}=\dfrac{\mathrm{Ze}}{8 \pi \epsilon_{0} \mathrm{r}^{2}} \quad$ using (1)
$\mathrm{E}=\dfrac{79 \times 1.6 \times 10^{-19}}{8 \times 3.14 \times 8.854 \times 10^{-12} \times 49 \times 10^{-30}}=1.16 \times 10^{21} \mathrm{NC}^{-1}$

So the correct answer is option B.

Note: Gauss law states that the electric flux through any closed surface is proportional to the total electric charge enclosed by this surface. According to Gauss's theorem the net-outward normal electric flux through any closed surface of any shape is equivalent to 1/ε0 times the total amount of charge contained within that surface. As it only works in certain situations (as does Ohm's Law), then it is not universal. A universal law will work in any situation. Well Gauss's law is one of Maxwell's equations, which are universal, whereas Coulomb's law is not.