Question

a.) What must the charge (sign and magnitude) of a 1.45 g particle be for it to remain balanced against gravity when placed in a downward-directed electric field of magnitude 650 N/C? Draw a diagram showing the particle, electric field, and forces.

b.) What is the magnitude of the electric field in which the electric force it exerts on a proton is equal in magnitude to the proton’s weight?

Mass of proton = 1.67262*10^-27 kg
Charge of proton = +1.60*10^-19 C

Answer 1

Concepts and reason

The concept required to solve the given problem is Newton’s second law and electric force.

Firstly, draw the free-body diagram of the system of charged particle and the gravity. Then, balance the force equation to find the magnitude and sign of the charge.

Then, again balance the force equation for the case of proton and find the magnitude of electric field.

Fundamentals

The expression of the weight of an object is,

$W = mg$

Here, W is the weight, m is the mass, and g is the acceleration due to gravity.

The force acting on a particle with charge q placed in an electric field is as follows:

$F = qE$

Here, E is the electric field.

(a)

The figure 1 explains the forces acting on the charged particle of mass m to balance it against the force of gravity.

The weight of an object always acts in downward direction. The given electric field is also in the downward direction.

The electric force must act in the upward direction for the system to maintain balance.

Thus, the charge on the particle must be negative.

The charge particle is balanced such that the net force acting on it is zero. The net force acting on the charged particle is as follows:

$\sum {{F_{{\rm{net}}}}} = F - W$

Substitute 0 for net force in the above expression.

$\begin{array}{c}\\F - W = 0\\\\F = W\\\end{array}$

Substitute qE for F and mg for W in the above expression and find the expression of q.

$\begin{array}{c}\\qE = mg\\\\q = \frac{{mg}}{E}\\\end{array}$

Substitute 1.45 g for m, $9.8{\rm{ m/}}{{\rm{s}}^{ - 2}}$ for g, and 650 N/C for E in the above expression.

$\begin{array}{c}\\q = \frac{{\left( {1.45{\rm{ g}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{650{\rm{ N/C}}}}\left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right)\\\\ = 2.2 \times {10^{ - 5}}{\rm{ C}}\\\end{array}$

(b)

The force on the proton is equal to the proton’s weight. The equation of force can be written as follows:

$F = W$

Substitute qE for F and mg for W in the above expression and find the expression of E.

$\begin{array}{c}\\qE = mg\\\\E = \frac{{mg}}{q}\\\end{array}$

Substitute $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for m, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for q in the above expression.

$\begin{array}{c}\\E = \frac{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\\\ = 1.02 \times {10^{ - 7}}{\rm{ N/C}}\\\end{array}$

Ans: Part a

The magnitude and sign of the charged particle are negative and $2.2 \times {10^{ - 5}}{\rm{ C}}$ respectively.