Question

A positive charge of 4.20 μCis fixed in place. From a distance of 3.90 cm a particle of mass5.00 g and charge +3.70 μC is fired with an initial speed of74.0 m/s directly toward the fixed charge. How close to the fixedcharge does the particle get before it comes to rest and startstraveling away?

Answer 1

Concepts and reason

The concept of conservation of energy is used here.

Initially, find the initial kinetic energy and potential energy when the particle of mass 5.0 g is fired with an initial speed ${v_1}$ . After that, find the final potential energy and equate both using the energy conservation law. Since, the particle having mass 5.0 g comes to rest. Therefore, the final kinetic energy is zero.

Substitute all the values in the equation and calculate the distance between both the charges when the particle having mass 5.0 g comes to rest.

Fundamentals

The law of conservation of energy states that sum of initial kinetic and potential energy is equal to the sum of final kinetic and potential energies. The total energy always remains conserved.

Mathematically, it is given as:

$K{E_i} + P{E_i} = K{E_f} + P{E_f}$

Here, i and f represents the initial and final states.

The expression of the kinetic energy of an object having mass m moving with velocity v is given as follows:

$KE = \frac{1}{2}m{v^2}$

The expression of the potential energy between the two charges separated by a distance R is given as follows:

$PE = \frac{{k{q_1}{q_2}}}{R}$

Here, k is the Coulomb’s constant, ${q_1}$ is the charge on first particle and ${q_2}$ is the charge on second particle.

The initial potential energy between the two charges separated by a distance R is given as:

$P{E_i} = \frac{{k{q_1}{q_2}}}{R}$

The initial kinetic energy of the particle having mass m moving with velocity ${v_1}$ is given as:

$K{E_i} = \frac{1}{2}m{v_1}^2$

The final potential energy between the two charges when the particle having mass m is at x from the first charge.

$P{E_f} = \frac{{k{q_1}{q_2}}}{x}$

The final kinetic energy of the particle having mass m is,

$K{E_f} = 0$

Since, the particle comes to rest finally when it is at x from the first charge.

Use the conservation of energy principle and substitute all the above equation in the expression $K{E_i} + P{E_i} = K{E_f} + P{E_f}$ .

$\begin{array}{c}\\\frac{1}{2}m{v_1}^2 + \frac{{k{q_1}{q_2}}}{R} = 0 + \frac{{k{q_1}{q_2}}}{x}\\\\\frac{1}{2}m{v_1}^2 = \frac{{k{q_1}{q_2}{\kern 1pt} }}{x} - \frac{{k{q_1}{q_2}}}{R}\\\\x = \frac{{k{q_1}{q_2}}}{{\left( {\frac{1}{2}m{v_1}^2 + \frac{{k{q_1}{q_2}}}{R}} \right)}}\\\end{array}$

Substitute $9.0 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}$ for k, $4.2{\rm{\mu C}}$ for ${q_1}$ , $3.7{\rm{ \mu C}}$ for ${q_2}$ , 5.0 g for m, 3.90 cm for R and 74.0 m/s for ${v_1}$ in equation $x = \frac{{k{q_1}{q_2}}}{{\left( {\frac{1}{2}m{v_1}^2 + \frac{{k{q_1}{q_2}}}{R}} \right)}}$ .

$\begin{array}{c}\\x = \frac{{\left( {9.0 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {4.2{\rm{\mu C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ \mu C}}}}} \right)} \right)\left( {{\rm{3}}{\rm{.7\mu C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ \mu C}}}}} \right)} \right)}}{{\left( \begin{array}{l}\\\frac{1}{2}\left( {5.0{\rm{ g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1.0{\rm{ g}}}}} \right)} \right){\left( {74.0{\rm{ m/s}}} \right)^2}\\\\ + \frac{{\left( {9.0 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {4.2{\rm{\mu C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ \mu C}}}}} \right)} \right)\left( {{\rm{3}}{\rm{.7\mu C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ \mu C}}}}} \right)} \right)}}{{3.90{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.0{\rm{ cm}}}}} \right)}}\\\end{array} \right)}}\\\\ = 0.0081{\rm{ m}}\\\end{array}$

Ans:

The distance between the fixed charge and the particle is 0.0081 m.