A gold nucleus ( Z = 79 ) has a radius of approximately 7.0 × 10^...

Given $Z = 79$, the total charge of the nucleus is $Q = Ze = 79 \times 1.6 \times 10^{-19} \text{ C}$.

The radius of the nucleus is $R = 7.0 \times 10^{-15} \text{ m}$.

The electric field inside a uniformly charged solid sphere at a distance $r$ from the center is given by:

$E = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q r}{R^3}$

At the middle point of a radius, $r = \dfrac{R}{2}$. Substituting this into the formula gives:

$E = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q \left(\dfrac{R}{2}\right)}{R^3} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{2 R^2}$

Substituting the known values:

$E = \dfrac{9 \times 10^9 \times 79 \times 1.6 \times 10^{-19}}{2 \times (7.0 \times 10^{-15})^2}$

$E = \dfrac{9 \times 126.4 \times 10^{-10}}{2 \times 49.0 \times 10^{-30}}$

$E = \dfrac{1137.6 \times 10^{-10}}{98.0 \times 10^{-30}}$

$E = 11.608 \times 10^{20} \text{ N C}^{-1}$

$E \approx 1.16 \times 10^{21} \text{ N C}^{-1}$