91影视

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 $v$.

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

$F=\frac{1}{4蟺{蔚}_0}\frac{q_1{q}_2}{R^2}\widehat{R}$ 鈥︹�� (1)

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

$F=\frac{1}{4蟺{蔚}_0}\frac{q_1{q}_2(1-\frac{v^2}{c^2})}{R^2}\widehat{R}$ 鈥︹�� (2)

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

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

$p={鈭�}_0鈭�(E脳B)d蟿$ 鈥︹�� (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.

$F_{12}+F_{21}$ 鈥︹�� (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 $q_1$.

(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).

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

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

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

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

(b)

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

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

$F_{21}=\frac{1}{4蟺{蔚}_0}\frac{q_1{q}_2(1鈭�\frac{v^2}{c^2})}{{\left(vt\right)}^2}$

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

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

$F_{21}=\frac{1}{4蟺{蔚}_0}\frac{q_1{q}_2(1鈭�\frac{v^2}{c^2})}{{\left(vt\right)}^2}\widehat{z}$

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

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

(c)

The expression for the linear momentum is given by,

$p={鈭�}_0鈭�(E脳B)d蟿$

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

$\begin{array}{l}p={鈭�}_0鈭�[(E_1+E_2)脳(B_1+B_2)]d蟿\\ p={鈭�}_0鈭�(E_1+B_2)d蟿\end{array}$ 鈥︹��. (5)

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

$E_1=\frac{1}{4蟺{蔚}_0}\frac{q_1}{r^2}\widehat{r}$

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

$E_2=\frac{q_2}{4蟺{蔚}_0}\frac{(1鈭�\frac{v^2}{c^2})}{{(1鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}\frac{\widehat{R}}{R^2}$

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

$E_2=\frac{q_2}{4蟺{蔚}_0}\frac{(1鈭�\frac{v^2}{c^2})}{{(1鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}\frac{(r鈭�vt)}{R^2}$

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

$B_2=\frac{1}{c^2}(v脳E_2)$

Substitute $\frac{q_2}{4蟺{蔚}_0}\frac{(1鈭�\frac{v^2}{c^2})}{{(1鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}\frac{(r鈭�vt)}{R^2}$for$E_2$into above equation.

$\begin{array}{l}{B}_2=\frac{1}{c^2}(v脳\frac{q_2}{4蟺{蔚}_0}\frac{(1鈭�\frac{v^2}{c^2})}{{(1鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}\frac{(r鈭�vt)}{R^2})\\ B_2=\frac{q_2(1鈭�\frac{v^2}{c^2})}{4蟺{蔚}_0{c}^2}\frac{v脳r}{{(R^2鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}\\ B_2=\frac{q_2(1鈭�\frac{v^2}{c^2})}{4蟺{蔚}_0{c}^2}\frac{vr\sin 胃}{{(R^2鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}\widehat{蠒}\end{array}$

Substitute $\frac{q_2(1鈭�\frac{v^2}{c^2})}{4蟺{蔚}_0{c}^2}\frac{vr\sin 胃}{{(R^2鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}\widehat{蠒}$for $B_2$and $\frac{1}{4蟺{蔚}_0}\frac{q_1}{r^2}\widehat{r}$ for $E_1$into equation (5).

$\begin{array}{l}p\left(t\right)={蔚}_0鈭�\frac{1}{4蟺{蔚}_0}\frac{q_1}{r^2}\widehat{r}脳\frac{q_2(1鈭�\frac{v^2}{c^2})}{4蟺{蔚}_0{c}^2}\frac{vr\sin 胃}{{(R^2鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}\widehat{蠒}\\ p\left(t\right)={蔚}_0\frac{q_1}{4蟺{蔚}_0}\frac{q_2(1鈭�\frac{v^2}{c^2})v}{4蟺{蔚}_0{c}^2}鈭�\frac{1}{r^2}\frac{r\sin 胃}{{(R^2鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}(\widehat{r}脳\widehat{蠒})\\ p\left(t\right)=\frac{q_1{q}_2(1鈭�\frac{v^2}{c^2})v}{{\left(4蟺c\right)}^2{蔚}_0}鈭�\frac{\sin 胃}{r{(R^2鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}(\widehat{r}脳\widehat{蠒})\end{array}$

Here,$\widehat{r}脳\widehat{蠒}=鈭�胃=鈭�(\cos 胃cos蠒\widehat{x}+\cos 蠒\sin 胃\widehat{y}鈭�\sin 胃\widehat{z})$, $x$and$y$component is integrated to zero.

$p\left(t\right)=\frac{q_1{q}_2(1鈭�\frac{v^2}{c^2})v}{{\left(4蟺c\right)}^2{蔚}_0}鈭�\frac{{\sin }^2胃}{r{(R^2鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}{r}^2\sin 胃drd胃d蠒$

Here,

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

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

$p\left(t\right)=\frac{q_1{q}_2(1鈭�\frac{v^2}{c^2})v}{8蟺c^2{蔚}_0}鈭�\frac{r{\sin }^3胃}{{(r^2+{\left(vt\right)}^2鈭�2vrt\cos 胃鈭�{\left(\frac{vr\sin 胃}{Rc}\right)}^2)}^{\frac{3}{2}}}drd蠒$

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

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

So, the linear momentum is,

$\begin{array}{l}p\left(t\right)=\frac{q_1{q}_2(1鈭�\frac{v^2}{c^2})v}{{蔚}_0}鈭�\frac{1}{vt{\sin }^2胃\left(\frac{1鈭�v^2}{c^2}\right)}[1+\frac{\cos 胃}{\sqrt{1鈭�{\left(\frac{v}{c}\right)}^2{\sin }^2胃}}]{\sin }^3胃d胃\\ p\left(t\right)=\frac{q_1{q}_2(1鈭�\frac{v^2}{c^2})v}{8蟺c^2{蔚}_0}\frac{1}{vt\left(\frac{1鈭�v^2}{c^2}\right)}鈭�\frac{1}{{\sin }^2胃}[1+\frac{\cos 胃}{\sqrt{1鈭�{\left(\frac{v}{c}\right)}^2{\sin }^2胃}}]{\sin }^3胃d胃\\ p\left(t\right)=\frac{q_1{q}_2}{8蟺c^2{蔚}_{0}t}{鈭�}_0^{蟺}\sin 胃d胃+\frac{c}{v}{鈭�}_0^{蟺}\frac{\cos 胃\sin 胃}{\sqrt{{\left(\frac{c}{v}\right)}^2鈭�{\sin }^2胃}}d胃\end{array}$

The integral value of${鈭�}_0^{蟺}\sin 胃d胃$is $2$ and second integral let$u=\cos 胃$so,

$du=鈭�\sin 胃d胃$

${鈭�}_0^{蟺}\frac{\cos 胃\sin 胃}{\sqrt{{\left(\frac{c}{v}\right)}^2鈭�{\sin }^2胃}}d胃={鈭�}_{鈭�1}^1\frac{u}{\sqrt{{\left(\frac{c}{v}\right)}^2鈭�1+u^2}}$

Since the integrand is odd and the interval is even.

${鈭�}_0^{蟺}\frac{\cos 胃\sin 胃}{\sqrt{{\left(\frac{c}{v}\right)}^2鈭�{\sin }^2胃}}d胃=0$

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

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

(d)

Calculate the sum of the forces.

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

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

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

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

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.