## HW2: Gauss Law

### 22.9

To compute The charge inside the box we apply Gauss' law to the two faces where the $\vec{E}$ field is perpendicular to the face (or parallel to the normal vector to the face's plane).

$\phi=(E_{left}-E_{right})*l^2)=(410N/m-540N/m)*625m^2)=\frac{Q_{in}}{\epsilon_0}$
$Q_{in}=-7.19*10^{-7} C$

### 22.10

This problem is very similar to the previous one, but in this case we know the charge inside
and we are asked to compute the flux through one of the faces. To do so we need to note that
all the faces will have the same flux through them, because the charge is at the center of the cubic box.
Therefore the $\phi_{face}=\phi_{total}/4=\frac{q}{\epsilon_0*4}$.

### 22. 21

Remember that the important concept in this problem is to know that inside a conductor
there cannot be electric fields. Therefore we have three different regions where the E field
will change. To solve these type of problems we need to apply Gauss' Law, but remembering that
when we do this for a surface within the conductor region, $\vec{E}=0$.

• a) 3cm from the center of the cavity, we are in vacuum with r=0.03 m: $\vec{E}=\frac{Q_{in}}{4r^2\pi\epsilon_0}=5.5*10^7 N/m$
• b) 6cm from the center we are inside the metal. Therefore E=0.
• c) 30 cm away from the center, we are again in the vacuum region, and all the charge inside is 5.5$\mu$C. Therefore, with r=0.3m $\vec{E}=\frac{Q_{in}}{4r^2\pi\epsilon_0}=5.5*10^5 N/m$.

### 22.22

This problem is like the previous one, but conceptual. We have a thin spherical conductor shell, i.e. a thin spherical metal shell.
There is a charge Q at the center of the sphere. In the recitation we showed how would the $\vec{E}$ field profile as a function of r b. This will be a function decaying as $1/r^2$ from r=0 to r=R$_{in}$. With R$_{in}$ being the internal radius of the shell. Then
from r=R$_{in}$ to r=R$_{out}$ E=0. With R$_{out}$ being the external radius of the shell. For r>R$_{out}$ again
$\vec{E}=\frac{Q_{in}}{4r^2\pi\epsilon_0}$.
The answer to question (a), which I assume asks if the shell affects the field away from the shell, is no. The field away from the shell is like the field we would have if the shell would not be there.
The answer to question B is yes. The free charge inside the conductor will rearrange itself in order to make the E field inside the shell=0. So the shell will be different if the charge would not be at the center.

### 22.31

This problem is a continuation of the previous two ones. We still have a metallic spherical shell, but now, this shell is charged,
with a charge Q. In addition, like in the previous two problems, there is a charge q at the center of the sphere. We are

• (a) What is the charge $\bf{q_{inner}}$ at the inner surface of the shell. To solve this problem we need to apply Gauss' law to a surface within the conductor. This surface encloses the inner surface of the shell (red dotted surface in the picture). The flux through this surface is zero. This is because the E field is zero inside the conductor, independently of its charge! Therefore: $\phi=0=Q_{in}/\epsilon_0=\frac{q+q_{inner}}{\epsilon_0}$ $\rightarrow$ $q_{inner}=-q$. The charge at the inner surface is minus the charge at the center.
• (b) What is the charge $\bf{q_{out}}$ at the outerer surface of the shell. To solve this problem we need to apply Gauss' law to a surface outside the conductor. This surface encloses the outer and the inner surface of the shell (green dotted surface in the picture). The flux through this surface is not zero, but $\frac{Q+q}{\epsilon_0}$, the total charge contained within this surface. Therefore, the charge in the conductor Q=$q_{inner}+q_{out}$ $\rightarrow$ $q_{out}=Q-q_{inner}=Q+q$. So if we sum all the charges (outer, inner and center) we should recover the total charge Q+q $\rightarrow$ $q_{out}+q_{inner}+q_{center}=Q+q + -q + q=Q+q$.
• c) The E field in the cavity r<$r_{inner}$ $E=\frac{q}{4\pi\epsilon_0 r^2}$.
• d) The E field within the conductor ($r_{inner}$<r<$r_{out}$ E=0. (it is a conductor!!).
• e) The E field outside the conductor r>$r_{out}$ $E=\frac{q+Q}{4\pi\epsilon_0 r^2}$.

### 22.32

This problem is like 22.31. I wont solve it. Everything is identical, you need to replace q by +/- Q in the formulas given before to answer question ©.

### 22.35

This problem also makes use of Gauss' law on a cylindrical geometry. Remember, the surface area of a cylinder of length l and radius R A=2$\pi$Rl. The density of charge $\sigma$=Q/A.

• a) For r<R1 E=0 because there is no charge in the vacuum region, only on the shells.
• b) For R1<r<R2 we apply Gauss' law choosing a cylindrical surface of radius r, between the two shells: $\phi=E*2\pi rl=\frac{Q}{\epsilon_0}$ Therefore $E=\frac{\sigma R_1}{\epsilon_0r}$.
• c) For r<R2 E=0 because any gaussian surface outside the second shell will enclose a zero total charge (-Q+Q=0).
• d) the kinetic energy $E_k$ of an electron orbiting at the intermediate distance between the two shells can be easily obtained. The centripetal force due to the circular motion of the electron must compensate the attractive force exerted by the charge in the inner shell. This force $F=q_{e}*E(\frac{R_1+R_2}{2})=q_{e}*\frac{2\sigma R1}{\epsilon_0(R_1+R_2)}$. The centripetal force of an electron orbiting along a circle of radius $r=\frac{R_1+R_2}{2}$ is $F_c=\frac{2 m_e*v^2}{R_1+R_2}$. Therefore we can extract the value of v by equaling these two expressions $F_e=F_c$. Therefore, $\bf{v^2}=\frac{q_e\sigma R_1}{m_e\epsilon_0}$. The kinetic energy (1/2m$v^2$) is $E_K=\frac{q_e\sigma R_1}{2\epsilon_0}$.

### 22.39

Now we have a solid non conductive sphere of radius r0 uniformly charged surrounded by a spherical conductive shell of inner and outer radius r1,r2. The density of charge in the non-conductor is $\rho_E$. To solve the E field inside the non-conductive sphere we apply Gauss' law to a spherical surface of radius r<r0.

• a) $E*4\pi r^2=\frac{\rho *4/3 \pi r^3}{\epsilon_0}$. Therefore $E=\frac{\rho r}{3\epsilon_0}$.
• b) $E=\frac{\rho*r_0^3}{3\epsilon_0 r^2}$
• c) E=0
• d) $E=\frac{(\rho*(4/3)*\pi*r_0^3)+Q}{4\pi\epsilon_0 r^2}$

### 22.65

• a) At the inner shell
• b) zero for both.