Question

A proton is suspended in the air by an electric field at the surface of Earth. What is the strength of this electric field?

Answer 1

Answer:

The strength of the electric field is $1.02\times10^{-7}\frac{N}{C}$

Explanation:

By Newton's first law $\sum\overrightarrow{F}=0$ tells us that to maintain suspended the proton (at rest), the net force on it should be zero, assuming there's only gravitational interaction between earth and proton we should apply a force in the opposite direction equal to the weight of the proton ($W=mg$, at earth surface) to make net force zero.

So, the electrostatic force on the charge should be:

$F_{e}=mg$ (1)

with m the mass and g the gravity

but electrostatic force of a charge on an electric field is:

$F_{e}=Eq$ (2)

with E the magnitude of electric field and q the charge of the proton

Equating (1) and (2):

$Eq=mg$

Solving for E:

$E=\frac{mg}{q}$

A fast search on google give us the mass of the proton and its charge:

$E=\frac{(1.67\times10^{-27}kg)(9.81\frac{m}{s^{2}})}{1.60\times10^{-19}C}=1.02\times10^{-7}\frac{N}{C}$