Question

A proton is released in a uniform electric field, and it experiences an electric force of 5 X 10^-10 N toward the South. What is the electric field the proton experiences?

Answer 1

Answer:

Electric field on proton

$E=3.12\times 10^9\ N/C$

Explanation:

Given that

$Force,F=5\times 10^{-10}\ N$

We know that

Charge on proton

$q=1.6\times 10^{-19}\ C$

We know that

Force = Electric field x Charge

F= E x q

$E=\dfrac{F}{q}\ N/C$

$E=\dfrac{5\times 10^{-10}}{1.6\times 10^{-19}}\ N/C$

$E=3.12\times 10^9\ N/C$

Electric field on proton

$E=3.12\times 10^9\ N/C$