HW 9

8-11

A proton has a positive charge of $e=1.6\times 10^{-19} ~\rm C$. So the force acting on it is ${\bf F}=e{\bf E} =- 9.6 \times 10^{-14}{\bf i}~\rm N$. (a) The acceleration is ${\bf a}= {\bf F}/m_p = -5.7\times 10^{13}{\bf i}~\rm m/s^2$. (b) Since work is done on the particle by $F\Delta x$ ( $\Delta x=7.00~\rm m$), it consumes the initial kinetic energy of the proton ${1\over 2}m_pv^2$. So $v=\sqrt{2F\Delta x /m_p}=2.84\times 10^7 ~\rm m/s$. (c) $\Delta t = v/a = 5.0\times 10^{-7}~\rm s$.

8-12

In order to stop an electron, the electric field has to be in the same direction as the direction of the electron motion. The force that takes it to stop is such that $F\Delta x = K$ where $K$ is the initial kinetic energy. So $F=1.60\times 10^{-16}~\rm N$. Then the magnitude of the electric field is $E= F/e = 1.00\times 10^3 ~\rm N/C$.

9-1

The area of the right face is in the $y$ direction and the magnitude is $1.4^2 = 1.96 ~\rm m^2$. So ${\bf A}= 1.96{\bf i}~\rm m^2$. The the total flux through this face is $\Phi ={\bf E}\cdot {\bf A}$. (a) ${\bf E}\cdot {\bf A}=0$. (b) ${\bf E}\cdot {\bf A}= -3.92 ~\rm Nm^2/C$. (c) Again, ${\bf E}\cdot {\bf A}=0$. (d) Since the flux going in the cube comes out as well, the total flux through the cube is identically zero.

9-2

Make a Gaussian surface which is cylindrical with the centre at the origin and with a radius $r$ and a height $L$. Then From Gauss' Law, $\Phi= 2\pi r LE = Q/\epsilon_0$. (a) For $r<a$, the Gaussian surface does not enclose any net charge inside. So the electric field is zero. (b) Between the two cylinders, the Gaussian surface encloses total charge of $Q= \lambda L$. Therefore, $E={ 1\over 2\pi\epsilon_0}{\lambda \over r}$.

9-3

Now make a Gaussian surface which encloses one of the metal plates, which gives a flux of $\Phi = Q/\epsilon_0$ where $Q$ is the total charge of one of the plates. But we know that the electric fields at outer surfaces is zero, neglecting the edge effects. That means $Phi = EA$ where $A$ is the inner surface area and $E$ is the electric fields at the inner surfaces. So $Q = \epsilon E A = 4.87\times 10^{-10} ~\rm C$.

9-4

Take a Gaussian surface which is spherical with a radius $r$. Then $Phi_g = 4\pi r^2 E_g$ where $E_g$ is the gravitational field. From the Gauss' Law, we get $E_g = -Gm/r^2$. The force is simply the gravitational field times the test mass. So $F_g(r) = -GmM/r^2$. The negative charge means that the force is always attractive.

9-5

It is simplest to get the total flux out of the cube: $\Phi = Q/\epsilon_0 = 1.92\times 10^7~\rm Nm^2/C$. The flux out of each face will be just $1/6$ of this total, since all the faces are identical. So $\Phi_f = 0.32\times 10^7 ~\rm Nm^2/C$. If the charge were not at the centre, the flux out of the faces will change, but the total flux remains the same.

9-6

Here we just need to set up Gaussian surfaces which is spherically symmetric with some radius $r$ and then count the charges enclosed by the surfaces. Then the resulting electric field is simply $E(r) ={1\over 4\pi \epsilon_0}{Q(r)/\over r^2}$. (a) At $r=0$, no charge is enclosed. So $E=0$. (b) $Q(r) = {4\pi\over 3}r^3 \rho$. So $E(r)= \rho r /3\epsilon_0$. At $r= 0.1 ~\rm m$, $E(r)=3.65\times 10^5 ~\rm N/C$. (c) At $r=40.0~\rm cm$, $E=14.6\times 10^5 ~\rm N/C$. (d) At $r=0.6~\rm m$, the total charge inside the Gaussian surface is the total charge of the sphere. So we just use the form $E= {1\over 4\pi \epsilon_0}{Q_{tot}/\over r^2}= 6.49\times 10^5 ~\rm N/C$.

9-7

You take a cylindrical Gaussian surface (again!) that encloses a circle segment of the wall of area $A$. Then the total charge enclosed is $Q= \sigma A$, whereas the total flux through the Gaussian surface would be $\Phi=2EA$. So Gauss' law says that $2EA= \sigma A/\epsilon_0$. So $E= \sigma /2\epsilon _0 = 0.48\times 10^6 ~\rm N/C$. Since the flux does not depend on the distance from the wall, the electric field does not have a distance dependence.

9-8

(a) With total charge $Q= 0.15 ~\rm C$ and the radius $r=0.15~\rm km$, the surface charge density is $\sigma = {Q\over 4\pi r^2}=5.3\times 10^{-7} ~\rm C/m^2$. (b) The electric field at the surface is $E= \sigma /\epsilon_0 = 6.0\times 10^4 ~\rm N/C$.