Question

EXAMPLE 4

Answer 1

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,

                                       $E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{x^{2}}$ V/m

               $\frac{1}{4\pi\epsilon_{0}}$ = 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