A charged cloud system produces an electric field in the air near Earth’s surface. A particle of charge $\mathbf{-}\mathbf{2}\mathbf{.0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{C}$is acted on by a downward electrostatic force of$\mathbf{3.0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{N}$when placed in this field. (a) What is the magnitude of the electric field? What are the (b) magnitude and (c) direction of the electrostatic Force${\overrightarrow{\mathbf{f}}}_{\mathbf{el}}$on the proton placed in this field? (d)What is the magnitude of the gravitational force${\overrightarrow{\mathbf{f}}}_{\mathbf{g}}$on the proton? (e) What is the ratio${\mathbf{F}}_{\mathbf{el}}\mathbf{}\mathbf{/}{\mathbf{F}}_{\mathbf{g}}$in this case?

• Charge of the particle,$\mathrm{q}=-2.0\times {10}^{-9}\text{C}$
• Downward electrostatic force,$\mathrm{F}=3.0\times {10}^{-6}\text{\hspace{0.17em}N}$
• Electric field is produced near the Earth’s surface.

The gravitational force on the particle is equal to the product of its mass and gravitational acceleration.

The electrostatic force is equal to the product of the charge and the electric field in which the charge is placed.

In this problem, we compare the strengths between the electrostatic force and the gravitational force.

Formulae:

The magnitude of the electrostatic force on a point charge,${\mathrm{F}}_{\mathrm{el}}=\mathrm{qE}$.

(i) Here, E is the magnitude of the electric field at the location of the particle.

The force of gravity on a particle, ${\mathrm{F}}_{\mathrm{g}}=\mathrm{mg}$. (ii)

The magnitude of the electric field strength is given asfollows using the given data in equation (i):

$\begin{array}{c}\mathrm{E}=\frac{\mathrm{F}}{\mathrm{q}}\\ =\frac{3.0\times {10}^{-6}\text{\hspace{0.17em}N}}{2.0\times {10}^{-9}\text{\hspace{0.17em}C}}\\ =1.5\times {10}^3\text{\hspace{0.17em}N/C}\end{array}$

Hence, the value of the electric field is$1.5\times {10}^3\text{\hspace{0.17em}N/C}$.

Since the force points downward and the charge is negative, the field $\overrightarrow{\mathrm{E}}$must point upward (in the opposite direction of $\overrightarrow{\mathrm{F}}$).

The magnitude of the electrostatic force on a proton using equation (i) and the charge oftheproton is given asfollows:

$\begin{array}{c}{\mathrm{F}}_{\mathrm{el}}=(1.60\times {10}^{-19}\text{\hspace{0.17em}C})(1.5\times {10}^3\text{\hspace{0.17em}N/C})\\ =2.4\times {10}^{-16}\text{\hspace{0.17em}N}\end{array}$

Hence, the value of the electrostatic force is $2.4\times {10}^{-16}\text{\hspace{0.17em}N}$.

A proton is positively charged, so the force is upward in the same direction as the field. Hence, the force is in the upward direction.

The magnitude of the gravitational force on the proton can be given using equation (ii) asfollows:

$\begin{array}{c}{\mathrm{F}}_{\mathrm{g}}=(1.67\times {10}^{-27}\text{\hspace{0.17em}kg})\left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\\ =1.6\times {10}^{-26}\text{\hspace{0.17em}N}\end{array}$

The force is downward. Hence, the value of the force is$1.6\times {10}^{-26}\text{\hspace{0.17em}N}$

The ratio of the forces can be given using part (b) and part (d) calculations:

$\begin{array}{c}\frac{{\mathrm{F}}_{\mathrm{el}}}{{\mathrm{F}}_{\mathrm{g}}}=\frac{2.4\times {10}^{-16}\text{\hspace{0.17em}N}}{1.64\times {10}^{-26}\text{\hspace{0.17em}N}}\\ =1.5\times {10}^{10}\end{array}$

The force of gravity on the proton is much smaller than the electrostatic force on the proton due to the field of strength. For the two forces to have equal strength, the electric field would have to be very small:

Hence, the ratio is $1.5\times {10}^{10}$.