Question

in a uniform electric field near the surface of earth, a particule having a charge of -2.0*10^-9C is acted on by a downward electric force of 3.0*10^-6N,

1. What is the magnitude and direction of the electric force exerted on a proton placed in the field?

2. What is the gravitational force on the proton?

3. What is the ratio of the electric force to the gravitational force in the case

Answer 1

1) Let's first find the magnitude and direction of the electric field:

$$F=qE,$$ $$E=\dfrac{F}{q}=\dfrac{3.0\times10^{-6}\ N}{-2.0\times10^{-9}\ C}=-1.5\times10^3\ \dfrac{N}{C}.$$

The magnitude of the electric field is $1.5\times10^3\ \dfrac{N}{C}$, direction is upward.

Then, we can find the magnitude and direction of the electric force exerted on a proton placed in the field:

$$F_{e,p}=q_pE=1.6\times10^{-19}\ C\times(-1.5\times10^3\ \dfrac{N}{C})=-2.4\times10^{-16}\ N.$$

The magnitude of the electric force exerted on a proton is $2.4\times10^{-16}\ N$. The direction is upward.

2) We can find the gravitation force on the proton as follows:

$$F_{g,p}=m_pg,$$ $$F_{g,p}=1.67\times10^{-27}\ kg\times9.8\ \dfrac{m}{s^2}=1.64\times10^{-26}\ N.$$

3) We can find the ratio of the electric force to the gravitational force as follows:

$$\dfrac{F_{e,p}}{F_{g,p}}=\dfrac{2.4\times10^{-16}\ N}{1.64\times10^{-26}\ N}=1.46\times10^{10}.$$