Answer:
Electric force, F = 20.16 N
Explanation:
It is given that,
Charge on gold nucleus, [tex]q_1=79e=79\times 1.6\times 10^{-19}=1.26\times 10^{-17}\ C[/tex]
An electron is sent at high speed toward a gold nucleus, [tex]q_2=1.6\times 10^{-19}\ C[/tex]
Distance between electron and gold nucleus, [tex]d=3\times 10^{-14}\ C[/tex]
We need to find the electric force acting on the electron. The formula for electrostatic force is given by :
[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times 1.26\times 10^{-17}\times 1.6\times 10^{-19}}{(3\times 10^{-14})^2}[/tex]
F = 20.16 N
The electric force acting on the electron is 20.16 N. Hence, this is the required solution.
