Question

What must the charge (sign and magnitude) of a particle of mass 1.46 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 630 N/C?
Use 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?
Use 1.67×10-27 kg for the mass of a proton, 1.60×10-19 C for the magnitude of the charge on an electron, and 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

Answer 1

Concepts and reason

The concepts required to answer the given question is Coulomb force, the electric field and the force due to gravity.

First calculate the sign of the charge of the particle. After that calculate the weight of the charge particle and divide it with the magnitude of electric field. the result will give charge on the particle. In second part first calculate the weight of the proton and then divide it with the charge of the proton. This will give the magnitude of electric field.

Fundamentals

The electric field at any point is defined as the coulombic force experienced by the unit positive charge when placed at that point. The direction of the electric field will be same as the force. It is denoted by $E$.

Weight of an object on earth is defined as the force exerted by the gravity of earth on that object. This can be expressed as,

$w = mg$.

Here$m$is the mass of that object and $g$is the acceleration due to gravity.

This is always in downward direction.

The equation of force in terms of charge $q$and electric field is expressed as,

$\vec F = q\vec E$

Here$q$ is charge on the particle and $\vec E$is electric field.

Since the weight always acts in downward direction then to remain stationary in electric field, the force acting on the particle must be in upward direction.

In the provided question, the electric field is in downward direction therefore to have upward force on the particle, charge on the particle must be negative.

Now weight of particle can be calculated by the equation as,

$w = mg$

And magnitude of force due to electric field can be calculated by the equation as,

$F = qE$

To remain in stationary position the weight and both forces should be equal. This can be written as,

$F = w$

Substitute $qE$for$F$and $mg$for$w$,

$qE = mg$

Rearrange above equation for $q$,

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

Substitute $1.46{\rm{ g}}$for$m$,$9.81{\rm{ m/}}{{\rm{s}}^2}$for$g$and$630{\rm{ N/C}}$for$E$in the above equation,

$\begin{array}{c}\\q = \frac{{\left( {1.46{\rm{ g}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {630{\rm{ N/C}}} \right)}}\\\\ = 22.7 \times {10^{ - 6}}{\rm{ C}}\\\end{array}$

Thus, charge on the particle is $- 22.7 \times {10^{ - 6}}{\rm{ C}}$

Weight of the proton can be calculated by the equation,

${W_p} = {m_p}g$

Here$m$ is mass of the proton and$g$ is acceleration due to gravity.

Force on proton due to electric field can be calculated as,

$F = qE$

Here, $q$ is charge on the proton and$E$ is magnitude of electric field.

As per the question the electric field is same as the weight of the proton.

It can be written as,

$qE = {m_p}g$

Rearrange above equation for $E$,

$E = \frac{{{m_p}g}}{q}$

Substitute $1.67 \times {10^{ - 27}}{\rm{ kg}}$for${m_p}$,$9.81{\rm{ m/}}{{\rm{s}}^2}$for$g$and $1.602 \times {10^{ - 19}}{\rm{ C}}$for$q$in the above equation.

$\begin{array}{c}\\E = \frac{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {1.602 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}\\\\ = 1.02 \times {10^{ - 7}}{\rm{ N/C}}\\\end{array}$

Ans:

The charge on the particle is $- 22.7 \times {10^{ - 6}}{\rm{ C}}$