Question

A proton is placed in a uniform electric field of 2950 N/C.
*Calculate the magnitude of the electric force felt by the proton.
(F = ? )
*Calculate the proton's acceleration.
( a= ? m/s² )

*Calculate the proton's speed after 1.40 ${\rm \mu s}$ in the field, assuming it starts from rest.

   ( V= ? m/s )

Answer 1

Concepts and reason

The concepts required to solve this question are electric field and Newton’s second law of motion.

First calculate the electric force felt by the proton due to the uniform magnetic field. After that, calculate the acceleration of the proton using the Newton’s second law of motion. Finally, calculate the speed of the proton using the kinematic equation of motion.

Fundamentals

The electric force experience by proton in uniform electric field is given by,

$F = qE$

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

Newton’s second law of motion states that the force acting on the particle is equal to the rate of change of momentum. That is, mass multiply by acceleration of the particle is the force applied on it. The expression for the force is,

$F = ma$

Here, m is the mass and $a$ is the acceleration of the particle.

The relation between the speed of the proton and acceleration is given by the kinematics equation as follows:

$v = u + at$

Here, u is the initial velocity, v is the final velocity, and t is the time taken.

The electric force experience by proton in uniform electric field is given by,

$F = qE$

Substitute $1.60 \times {10^{ - 19}}{\rm{ C}}$ for q and 2950 N/C for E in above equation.

$\begin{array}{c}\\F = \left( {1.60 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {2950{\rm{ N/C}}} \right)\\\\ = 4.72 \times {10^{ - 16}}{\rm{ N}}\\\end{array}$

Rearrange the expression $F = ma$ for the acceleration.

$a = \frac{F}{m}$

Substitute $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for m and $4.72 \times {10^{ - 16}}{\rm{ N}}$ for F.

$\begin{array}{c}\\a = \frac{{4.72 \times {{10}^{ - 16}}{\rm{ N}}}}{{1.67 \times {{10}^{ - 27}}{\rm{ kg}}}}\\\\ = 2.826 \times {10^{11}}{\rm{ m/}}{{\rm{s}}^2}\\\\ \approx 2.83 \times {10^{11}}{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The proton starts from rest. Thus, the initial velocity of the proton is zero. The relation between the speed of the proton and acceleration is given by the kinematics equation as follows:

$v = u + at$

Substitute 0 m/s for $u$ , $2.83 \times {10^{11}}{\rm{ m/}}{{\rm{s}}^2}$ for v, and $1.40{\rm{ }}\mu {\rm{s}}$ for t.

$\begin{array}{c}\\v = 0{\rm{ m/s}} + \left( {2.83 \times {{10}^{11}}\,{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.40{\rm{ }}\mu {\rm{s}}} \right)\left( {\frac{{{{10}^{ - 6}}\,{\rm{s}}}}{{1{\rm{ }}\mu {\rm{s}}}}} \right)\\\\ = 3.96 \times {10^5}{\rm{ m/s}}\\\end{array}$

Ans:

The magnitude of the electric force is $4.72 \times {10^{ - 16}}{\rm{ N}}$ .