Chapter 24, Problem 18P

(a)

To determine

The velocity of insulating charged sphere after collide.

Answer to Problem 18P

The velocity of insulating sphere 1 after collide is $\sqrt{k_{\text{e}}\frac{2m_{\text{2}}{q}_{\text{1}}{q}_{\text{2}}\left(\frac{1}{d}+\frac{1}{r_{\text{1}}+r_{\text{2}}}\right)}{m_{\text{1}}\left(m_1+m_2\right)}}$ and the velocity of insulating sphere 2 after collide is $\sqrt{k_{\text{e}}\frac{2m_1{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)}{m_2\left(m_1+m_2\right)}}$.

Explanation of Solution

Given info: The radius of sphere 1 is $r_1$, the radius of sphere 2 is $r_2$, the mass of sphere 1 is $m_1$, the mass of sphere 2 is $m_2$, the charge on sphere 1 $q_1$, the charge on sphere 2 is $q_2$, the distance between both charged insulated sphere is $d$.

Consider the diagram of two insulating sphere having charge $q_1$ and $q_2$ is given below,

Figure (1)

Write the expression to calculate the electric potential energy before collide,

$\Delta U_{{\text{e}}_{\text{1}}}=k_{\text{e}}\frac{q_1{q}_{\text{2}}}{d}$ (1)

Here,

$q_1$ is the charge on insulation sphere 1.

$q_2$ is the charge on insulating sphere 2.

$d$ is the distance between insulating sphere.

Write the expression to calculate the potential energy after collide,

$\Delta U_{\text{e}}=k_{\text{e}}\frac{q_1{q}_2}{r_1+r_2}$ (2)

Here,

$r_1$ is the radius of insulating sphere 1.

$r_2$ is the radius of the insulating sphere 2.

Add equation (1) and equation (2).

$\begin{array}{c}\Delta U_{\text{e}}=\Delta U_{{\text{e}}_{\text{1}}}+\Delta U_{{\text{e}}_2}\\ \Delta U_{\text{e}}=k_{\text{e}}\frac{q_1{q}_2}{d}+k_{\text{e}}\frac{q_1{q}_2}{r_1+r_2}\\ =k_{\text{e}}\left(\frac{q_1{q}_2}{d}+\frac{q_1{q}_2}{r_1+r_2}\right)\\ =k_{\text{e}}{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)\end{array}$

Write the equation of kinetic energy stored in charged sphere after collide, if velocity of charged sphere 1 is $v_1$ and velocity of charged sphere 2 is $v_2$.

$\Delta U_{\text{k}}=\frac{1}{2}\left(m_1{v}_1{}^2+m_2{v}_2{}^2\right)$

Here,

$m_1$ is the mass of insulating sphere 1.

$m_2$ is the mass of insulating sphere 2.

$v_1$ is the velocity of insulating sphere 1.

$v_2$ is the velocity of the insulating sphere 2.

From the law of conservation, both charged mass gets kinetic energy on diminishing of electric potential energy.

Then for equilibrium condition both energies will be same that is,

$-\Delta U_{\text{e}}=\Delta U_{\text{k}}$ (3)

The negative sign shows that there is decrease in electric potential energy.

Substitute $k_{\text{e}}{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)$ for $\Delta U_{\text{e}}$ and $\frac{1}{2}\left(m_1{v}_1{}^2+m_2{v}_2{}^2\right)$ for $\Delta U_{\text{k}}$ in equation (3).

$k_{\text{e}}{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)=\frac{1}{2}\left(m_1{v}_1{}^2+m_2{v}_2{}^2\right)$ (4)

Write the equation for conservation of momentum for final velocities of charged spheres.

$\begin{array}{c}{m}_1{v}_1=m_2{v}_2\\ v_1=\frac{m_2{v}_2}{m_1}\end{array}$ (5)

Substitute $\frac{m_2{v}_2}{m_1}$ for $v_1$ in equation (4).

$\begin{array}{c}{k}_{\text{e}}{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)=\frac{1}{2}\left(m_1{\left(\frac{m_2{v}_2}{m_1}\right)}_1{}^2+m_2{v}_2{}^2\right)\\ k_{\text{e}}{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)=\frac{1}{2}{m}_2{v}_2{}^2\left(\frac{m_2}{m_1}+1\right)\\ k_{\text{e}}{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)=\frac{1}{2}\left(\frac{m_2}{m_1}\right)\left(m_1+m_2\right)v_2{}^2\\ v_2=\sqrt{k_{\text{e}}\frac{2m_1{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)}{m_2\left(m_1+m_2\right)}}\end{array}$

Substitute $\sqrt{k_{\text{e}}\frac{2m_1{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)}{m_2\left(m_1+m_2\right)}}$ for $v_2$ in equation (5).

$\begin{array}{c}{v}_1=\frac{m_2}{m_1}\left(\sqrt{k_{\text{e}}\frac{2m_1{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)}{m_2\left(m_1+m_2\right)}}\right)\\ =\sqrt{k_{\text{e}}\frac{2m_2{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)}{m_1\left(m_1+m_2\right)}}\end{array}$

Conclusion:

Therefore, the velocity of sphere 1 after collide is $\sqrt{k_{\text{e}}\frac{2m_2{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)}{m_1\left(m_1+m_2\right)}}$ and the velocity of charged sphere 2 after collide is $\sqrt{k_{\text{e}}\frac{2m_1{q}_1{q}_2\left(\frac{1}{d}+\frac{1}{r_1+r_2}\right)}{m_2\left(m_1+m_2\right)}}$.

(b)

To determine

The effect on velocity of spheres after collide if the sphere were conductor.

Answer to Problem 18P

The velocity of sphere will be greater for conducting sphere than insulating sphere.

Explanation of Solution

Given info: The radius of sphere 1 is $r_1$, the radius of sphere 2 is $r_2$, the mass of sphere 1 is $m_1$, the mass of sphere 2 is $m_2$, the charge on sphere 1 $q_1$, the charge on sphere 2 is $q_2$, the distance between both charged insulated sphere is $d$.

Due to polarization, the most of the positive charge of one sphere at the centre and most of the negative charge at the centre of other sphere will attracts each other due to which their average centre distance will be less then geometric centre distance. Hence potential energy will be less and kinetic energy will be more for conducting sphere, hence due to more kinetic energy velocities of conducting spheres after collide will be more.

Conclusion:

Therefore, the velocities of conducting sphere after collide will be more than velocities of insulating sphere after collide due to effect polarization.