Answer:
E = 1.14 x 10⁹ V/m
Explanation:
Given,
The atomic number of gold, A = 79
The number of protons present in the nucleus, p = 79
The charge of a proton, q = 1.602 x 10⁻¹⁹ C
Therefore the total charge is, Q = 79 x 1.602 x 10⁻¹⁹
= 1.26558 x 10⁻¹⁷ C
The distant from the point charge, x = 10 nm
The electric field strength at a distant x is given by,
[tex]E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{x^{2}}[/tex] V/m
[tex]\frac{1}{4\pi\epsilon_{0}}[/tex] = 9 x 10⁹ Nm²C⁻²
Substituting the values in the above equation
E = (9 x 10⁹ X 1.26558 x 10⁻¹⁷) / (10 x 10⁻⁹)²
E = 1.14 x 10⁹ V/m
Hence, the electric field strength is, E = 1.14 x 10⁹ V/m
