HW 8

8-1

(a) Using the Coulomb's law, $F=e^2/(4\pi\epsilon r^2)=1.59\times 10^{-9}~\rm N$. (b) $Gm_p^2/r^2 = 1.29\times 10^{-45}~\rm N$. (c) By comparison, $Gm^2/r^2 =q^2/(4\pi\epsilon r^2)$, we obtain $q/m = \sqrt{4\pi \epsilon G} =8.61\times 10^{-11} ~\rm C/kg$. Compare this with that of an electron $e/m_e = 1.76\times 10^{11}~\rm C/kg$.

8-2

(a) Using the Coulomb's law, $F=e^2/(4\pi\epsilon r^2)=8.22\times 10^{-8}~\rm N$. (b) The centripetal force $F_c = mv^2/r$. If you equate this with the Coulomb force, $v=\sqrt{Fr/m_e}=2.18\times 10^{6} ~\rm m/s$.

8-3

(a) Using the Coulomb's law, $F=q_1q_2/(4\pi\epsilon r^2)= 1.6~\rm N$. (b) The magnitude of force between $q_1$ and $q_2$ is the same as that between $q_1$ and $q_3$. The direction of the two forces make an angle of $60~^{\rm o}$. So the total force is $F_{tot} = \sqrt{3} F =2.8 ~\rm N$.

8-4

Since the force acting on $q_3$ is zero, the electric field there is also zero. If you add the electric fields, $E_{tot}\propto q_1/(2d)^2 +q_2/d^2 = 0$. Then $q_1=-4q_2$.

8-5

The electric field is the sum of that of the upper cloud and that of the lower cloud: ${\bf E}= {1\over 4\pi \epsilon}(\hat{\bf z}q_1/r_1^2-\hat{\bf z}q_2 /r_2^2) =7.2\times 10^5 ~\rm N/C$ directed towards the ground.

8-6

In terms of the $(x,y)$ coordinate, the electric field due to the first charge is $E_0(-1/\sqrt{1.25}, 0.5/\sqrt{1.25})$, and that due to the second charge is $E_0(1/\sqrt{1.25}, 0.5/\sqrt{1.25})$, where $E_0 = q/(4\pi\epsilon r^2)=2.88\times 10^{-2} ~\rm N/C$, since the distance is $r= \sqrt{1.25} ~\rm m$. If you add the two electric fields, you get $E_{tot}= E_0/\sqrt{1.25}= 2.57\times 10^{-2} ~\rm N/C$ in the $y$ direction. (b) The electric force is $qE_{tot} = 7.71\times 10^{-8}~\rm N$ in the negative $y$ direction.

8-7

The dipole moment is defined as ${\bf p} = qd\hat{\bf z}$. An alternative form of the electric field is ${\bf E}= {1\over 4\pi \epsilon} {q{\bf r}/r^3}$, which we will use for this problem. The field due to the lower charge is ${\bf E}_1 = {-q\over 4\pi \epsilon (r^2+d^2/4)^{3/2} }~(r, d/2)$ in terms of the $(x,y)$ coordinate, and that due to the upper charge is ${\bf E}_2 = {q\over 4\pi \epsilon (r^2+d^2/4)^{3/2} }~(r, -d/2)$. The total field is ${\bf E}_{tot}= -{qd\over 4\pi \epsilon (r^2+d^2/4)^{3/2} }~(0,1)\approx -{{\bf p}\over 4\pi \epsilon r^3}$.

8-8

Firstly, the direction of the electric field would be in the negative $x$ direction at the origin. In order to sum up the electric field due to a segment $dx$ of line charge at $x$, we should first know the segment of the electric field, which is $dE = {1\over 4\pi\epsilon } {\lambda_0 dx\over x^2}$. Then we can integrate this over the range $(x_0,\infty )$. So $\int dE = {\lambda_0\over 4\pi\epsilon x_0}$, in the $-x$ direction.

8-9

From the formula $E(z) = {\sigma \over 2\epsilon}( 1-{z\over \sqrt{z^2+R^2} })$, you use the approximation ${z\over \sqrt{z^2+R^2} } \approx 1-{R^2\over 2z^2}$. Then $E(z)\approx {\sigma \pi R^2 \over 4\pi\epsilon z^2}$.

8-10

Use your common sense in this problem.