Question

A particle of charge $q_1$is at rest at the origin. A second particle, of charge$q_2$ , moves along the axis at constant velocity $v$.

(a) Find the force $F_{12}\left(t\right)$ of${\mathit{q}}_{\mathbf{1}}$ on $q_2$, at time$t$ . (When$q_2$ is at $z=vt$).

(b) Find the force $F_{21}\left(t\right)$of$q_2$ on$q_1$ , at time $t$. Does Newton's third law hold, in this case?

(c) Calculate the linear momentum$p\left(t\right)$ in the electromagnetic fields, at time$t$ . (Don't bother with any terms that are constant in time, since you won't need them in part (d)). [Answer:$({\mu }_0{q}_1{q}_2/4\pi t)$ ]

(d) Show that the sum of the forces is equal to minus the rate of change of the momentum in the fields, and interpret this result physically.

Short answer

(a) The force due to charge$q_1$ on charge$q_2$ at time tis $F_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(vt\right)}^2}\widehat{z}$.

(b) The force due to charge$q_2$ on charge$q_1$ at time t is$F_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2(1-\frac{v^2}{c^2})}{{\left(vt\right)}^2}\widehat{z}$ and the Newton’s third law does not hold.

(c) The expression for the linear momentum is $\frac{{\mu }_0{q}_1{q}_2}{4\pi t}\widehat{z}$.

(d) The expression for the sum of the forces is $\frac{{\mu }_0}{4\pi }\frac{q_1{q}_2}{t^2}\widehat{z}$.

Step-by-step solution

Write the given data from the question.

The charge $q_1$is at rest and charge$q_2$ is moving along $z$-axis.

The constant speed of the charge$q_2$ is $v$.

Determine the formulas to calculate the forces due to q1 and q2,the linear moment and sum of forces.

The expression to calculate the force between the two charges is given as follows.

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{R^2}\widehat{R}$ …… (1)

The expression to calculate the force due to moving charge is given as follows.

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2(1-\frac{v^2}{c^2})}{R^2}\widehat{R}$ …… (2)

Here,$c$is the velocity of the light.

The expression to calculate the linear momentum is given as follows.

$p={\in }_0\int (E\times B)d\tau$ …… (3)

Here,$E$is the electric field and $B$is the magnetic field.

The expression to calculate the sum of the forces is given as follows.

$F_{12}+F_{21}$ …… (4)

Here, is the force due to charge to charge and is the force due to charge to .

Here, $F_{12}$is the force due to charge $q_1$to charge $q_2$and$F_{21}$ is the force due to charge $q_2$to $q_1$.

Determine the force due to charge due to q1 on q2.

(a)

Consider the figure which represents the charge $q_1$ is on rest and charge $q_2$is moving along the $z$axis.

Calculate the force due to charge$q_1$on charge$q_2$at time t.

Substitute $vt$for$R$into equation (1).

$F_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(vt\right)}^2}\widehat{R}$

Substitute$\widehat{z}$ for $\widehat{R}$into above equation.

$F_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(vt\right)}^2}\widehat{z}$

Hence the force due to charge$q_1$on charge $q_2$at time tis $F_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(vt\right)}^2}\widehat{z}$.

Determine the force dur to charge q2 on charge q1at time t.

(b)

Calculate the force due to charge on charge at time t.

Substitute $vt$for$R$into equation (2).

$F_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2(1-\frac{v^2}{c^2})}{{\left(vt\right)}^2}$

The charge$q_2$is moving along the$z$axis.

Substitute $-\widehat{z}$for $\widehat{R}$into above equation.

$F_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2(1-\frac{v^2}{c^2})}{{\left(vt\right)}^2}\widehat{z}$

The Newton’s third law does not hold, because the vector of $(1-\frac{v^2}{c^2})$.

Hence the force due to charge$q_2$ on charge $q_1$at time tis $F_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2(1-\frac{v^2}{c^2})}{{\left(vt\right)}^2}\widehat{z}$.

Calculate the linear momentum.

(c)

The expression for the linear momentum is given by,

$p={\in }_0\int (E\times B)d\tau$

Substitute$E_1+E_2$for $E$and $B_1=B_2$for$B$into above equation.

$\begin{array}{l}p={\in }_0\int [(E_1+E_2)\times (B_1+B_2)]d\tau \\ p={\in }_0\int (E_1+B_2)d\tau \end{array}$ ……. (5)

The expression for the electric field due to charge$q_1$is given by,

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r^2}\widehat{r}$

The expression for the electric field due to charge $q_2$is given by,

$E_2=\frac{q_2}{4\pi {\epsilon }_0}\frac{(1-\frac{v^2}{c^2})}{{(1-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}\frac{\widehat{R}}{R^2}$

Substitute$r-vt$for $\widehat{R}$into above equation.

$E_2=\frac{q_2}{4\pi {\epsilon }_0}\frac{(1-\frac{v^2}{c^2})}{{(1-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}\frac{(r-vt)}{R^2}$

The expression for the magnetic field due to charge$q_2$is given by,

$B_2=\frac{1}{c^2}(v\times E_2)$

Substitute $\frac{q_2}{4\pi {\epsilon }_0}\frac{(1-\frac{v^2}{c^2})}{{(1-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}\frac{(r-vt)}{R^2}$for$E_2$into above equation.

$\begin{array}{l}{B}_2=\frac{1}{c^2}(v\times \frac{q_2}{4\pi {\epsilon }_0}\frac{(1-\frac{v^2}{c^2})}{{(1-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}\frac{(r-vt)}{R^2})\\ B_2=\frac{q_2(1-\frac{v^2}{c^2})}{4\pi {\epsilon }_0{c}^2}\frac{v\times r}{{(R^2-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}\\ B_2=\frac{q_2(1-\frac{v^2}{c^2})}{4\pi {\epsilon }_0{c}^2}\frac{vr\sin \theta }{{(R^2-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}\widehat{\varphi }\end{array}$

Substitute $\frac{q_2(1-\frac{v^2}{c^2})}{4\pi {\epsilon }_0{c}^2}\frac{vr\sin \theta }{{(R^2-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}\widehat{\varphi }$for $B_2$and $\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r^2}\widehat{r}$ for $E_1$into equation (5).

$\begin{array}{l}p\left(t\right)={\epsilon }_0\int \frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r^2}\widehat{r}\times \frac{q_2(1-\frac{v^2}{c^2})}{4\pi {\epsilon }_0{c}^2}\frac{vr\sin \theta }{{(R^2-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}\widehat{\varphi }\\ p\left(t\right)={\epsilon }_0\frac{q_1}{4\pi {\epsilon }_0}\frac{q_2(1-\frac{v^2}{c^2})v}{4\pi {\epsilon }_0{c}^2}\int \frac{1}{r^2}\frac{r\sin \theta }{{(R^2-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}(\widehat{r}\times \widehat{\varphi })\\ p\left(t\right)=\frac{q_1{q}_2(1-\frac{v^2}{c^2})v}{{\left(4\pi c\right)}^2{\epsilon }_0}\int \frac{\sin \theta }{r{(R^2-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}(\widehat{r}\times \widehat{\varphi })\end{array}$

Here,$\widehat{r}\times \widehat{\varphi }=-\theta =-(\cos \theta cos\varphi \widehat{x}+\cos \varphi \sin \theta \widehat{y}-\sin \theta \widehat{z})$, $x$and$y$component is integrated to zero.

$p\left(t\right)=\frac{q_1{q}_2(1-\frac{v^2}{c^2})v}{{\left(4\pi c\right)}^2{\epsilon }_0}\int \frac{{\sin }^2\theta }{r{(R^2-{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}{r}^2\sin \theta drd\theta d\varphi$

Here,

$\begin{array}{l}R=(r-vt)\\ R^2=r^2+{\left(vt\right)}^2-2vrt\cos \theta \end{array}$

Substitute$r^2+{\left(vt\right)}^2-2vrt\cos \theta$for $R^2$into above equation.

$p\left(t\right)=\frac{q_1{q}_2(1-\frac{v^2}{c^2})v}{8\pi c^2{\epsilon }_0}\int \frac{r{\sin }^3\theta }{{(r^2+{\left(vt\right)}^2-2vrt\cos \theta -{\left(\frac{vr\sin \theta }{Rc}\right)}^2)}^{\frac{3}{2}}}drd\varphi$

Integrate the $r$integral according to the CRC table,

$\begin{array}{l}=\frac{2}{\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }[2vt\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }-2vt\cos \theta ]}\\ =\frac{1}{vt\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }[\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }-\cos \theta ]}\\ =\frac{[\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }+\cos \theta ]}{vt\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }[1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta -{\cos }^2\theta ]}\\ =\frac{1}{vt{\sin }^2\theta \left(\frac{1-v^2}{c^2}\right)}[1+\frac{\cos \theta }{\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }}]\end{array}$

So, the linear momentum is,

$\begin{array}{l}p\left(t\right)=\frac{q_1{q}_2(1-\frac{v^2}{c^2})v}{{\epsilon }_0}\int \frac{1}{vt{\sin }^2\theta \left(\frac{1-v^2}{c^2}\right)}[1+\frac{\cos \theta }{\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }}]{\sin }^3\theta d\theta \\ p\left(t\right)=\frac{q_1{q}_2(1-\frac{v^2}{c^2})v}{8\pi c^2{\epsilon }_0}\frac{1}{vt\left(\frac{1-v^2}{c^2}\right)}\int \frac{1}{{\sin }^2\theta }[1+\frac{\cos \theta }{\sqrt{1-{\left(\frac{v}{c}\right)}^2{\sin }^2\theta }}]{\sin }^3\theta d\theta \\ p\left(t\right)=\frac{q_1{q}_2}{8\pi c^2{\epsilon }_{0}t}{\int }_0^{\pi }\sin \theta d\theta +\frac{c}{v}{\int }_0^{\pi }\frac{\cos \theta \sin \theta }{\sqrt{{\left(\frac{c}{v}\right)}^2-{\sin }^2\theta }}d\theta \end{array}$

The integral value of${\int }_0^{\pi }\sin \theta d\theta$is $2$ and second integral let$u=\cos \theta$so,

$du=-\sin \theta d\theta$

${\int }_0^{\pi }\frac{\cos \theta \sin \theta }{\sqrt{{\left(\frac{c}{v}\right)}^2-{\sin }^2\theta }}d\theta ={\int }_{-1}^1\frac{u}{\sqrt{{\left(\frac{c}{v}\right)}^2-1+u^2}}$

Since the integrand is odd and the interval is even.

${\int }_0^{\pi }\frac{\cos \theta \sin \theta }{\sqrt{{\left(\frac{c}{v}\right)}^2-{\sin }^2\theta }}d\theta =0$

Thus,$p\left(t\right)=\frac{{\mu }_0{q}_1{q}_2}{4\pi t}\widehat{z}$

Hence the expression for the linear momentum is $\frac{{\mu }_0{q}_1{q}_2}{4\pi t}\widehat{z}$

Calculate the expression for the sum of the forces.

(d)

Calculate the sum of the forces.

Substitute $\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(vt\right)}^2}\widehat{z}$for $F_{12}$and $\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2(1-\frac{v^2}{c^2})}{{\left(vt\right)}^2}\widehat{z}$ for $F_{21}$into equation (4).

$\begin{array}{l}{F}_{12}+F_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(vt\right)}^2}\widehat{z}-\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2(1-\frac{v^2}{c^2})}{{\left(vt\right)}^2}\widehat{z}\\ F_{12}+F_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{v^2{t}^2}(1-1+\frac{v^2}{c^2})\widehat{z}\\ F_{12}+F_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{c^2{t}^2}\widehat{z}\end{array}$

Substitute${\mu }_0{\epsilon }_0$for $\frac{1}{c^2}$into above equation.

$\begin{array}{l}{F}_{12}+F_{21}=\frac{{\mu }_0{\epsilon }_o}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{t^2}\widehat{z}\\ F_{12}+F_{21}=\frac{{\mu }_0}{4\pi }\frac{q_1{q}_2}{t^2}\widehat{z}\end{array}$

Since $q_1$is at the rest and $q_2$is moving, there should be exist an equivalent force $F_{mech}$which balance the sum of the forces. We have found$\frac{dp}{dt}=F_{mech}$ , that means that impulse imparted to the system by the external force ends up as momentum in the field.