## HW3: Chapter 23, Electrostatic Potential

2,3,14,16,19,34,36,61,71,84

### 23.2

Note that my numbers might not match those in the book!

How much work does the electric field do in moving a proton from a point with a potential of +110 V to a point where it is -25 V? Express your answer both in joules and electron volts.

This is an easy problem: Work is minus the electrostatic potential energy. The Elec. Potential energy is given by the potential times the charge. We are moving a proton from a high potential region to a low potential region. This proton is losing electrostatic potential energy, which means that it is gaining kinetic energy. The proton is being accelerated by the electric field. Protons move in the direction of the electric field, therefore slide down the electrostatic potential ramp. W$_{a \rightarrow b}$=-$\Delta V_{a \rightarrow b} * q$.
$\Delta V_{a,b}=V_b -V_a$=-25-110=-135 V
W=135V*1.6*10$^{-19}$ C=2.16e-17 J=135 eV

### 23.3

Note that my numbers might not match those in the book!

An electron acquires 15.8 × 10-16 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference $\Delta V_{A\rightarrow B}$ between the plates?

This is the continuation of the previous problem, but now we have an electron instead of a proton, so we have to reverse signs. Electrons move against $\vec{E}$. This means that they climb electrostatic potential ramps. The $V_B$ must be larger than $V_A$, $\Delta V_{A \rightarrow B}>0$ . But the electron loses potential energy on going from A to B because its charge is negative: $\Delta U_{A \rightarrow B}=q_e * \Delta V_{A \rightarrow B}$. As all the potential energy lost by the electron is converted into kinetic energy we have:
$\Delta V_{A \rightarrow B}=-\frac{\Delta U_{A \rightarrow B}}{q_e}$=9880 V

Because $\Delta V_{A\rightarrow B}>0$, plate B is at a higher potential.

### 23.14

Note that my numbers might not match those in the book!

A 32 cm diameter conducting sphere is charged to 500 V relative to V = 0 at r =$\infty$. What is the surface charge density $\sigma$?

From Gauss' law we know that $E_{inside}=0$ (conductor) and $E_{out}=\frac{Q}{4\pi\epsilon_0 r^2}$.
$V=-\int{E dr}$. $V_{out}=\frac{Q}{4\pi\epsilon_0 r}$.
$V_{in}=constant$.

At r=R=0.32 m, 500 V=$\frac{Q*9*10^9 V*m}{0.32 m}$. $\sigma=\frac{Q}{4\pi R^2}$=2.77 *10$^{-8}$ C/m$^2$.

At what distance will the potential due to the sphere be only 20 V?
$r=\frac{Q*9*10^9 V*m}{20 V}$=4 m.

### 23.16

Very long wire with uniform charge per unit lenght, $\lambda$. Determine the potential difference between two points Ra and Rb

This is one of those cases where we obtain V from E. And we obtain E by using Gauss' law. We do not need to know any formula to solve this problem. We just need to know Gauss' law and that $V_{a,b}=-\int_a^b{\vec{E(r)} dr}$.

The E field as a function of r for a very long wire, length l is obtained from Gauss' law:
$Q=\lambda*l$
$E*2\pi rl=\frac{Q_{in}}{\epsilon_0}=\frac{\lambda*l}{\epsilon_0}\rightarrow E(r)=\frac{\lambda}{2\pi\epsilon_0 r}$.

Therefore:

$V_{a,b}=-\int_a^b{\vec{E(r)} dr}=-\int_a^b{\frac{\lambda}{2\pi\epsilon_0 r} dr}=\frac{\lambda}{2\pi\epsilon_0}ln\frac{R_a}{R_b}$

### 23.19

As in the previous problem we use Gauss' law to determine E and then we obtain V by integrating E.

We have a nonconducting sphere, with a total charge Q uniformly distributed throughout its volume. The E field inside the sphere is not zero, because it is not a conductor and there is charge everywhere inside. We define to regions: r<r0 (inside) and r>ro (outside).
We also make use of the charge density instead of the total charge: $\sigma=\frac{Q}{4/3\pi r_0^3}$
Inside, r<r0:

$E=\frac{\sigma r}{3\epsilon_0}$
$V=\frac{\sigma r^2}{6\epsilon_0}+constant$

Outside, r>r0:

$E=\frac{Q}{4\pi\epsilon_0 r^2}$
$V=\frac{Q}{4\pi\epsilon_0 r}$

We can determine the value of the constant by making the two expressions for V equal at r=r0.

### 23.34

The following set of problems involve computing the potential as the sum of the potential generated by a discrete charge distribution. In this case we do not use Gauss' law to obtain E, but we directly sum or integrate the charge distribution

For this problem we have a square of side l. qe have 3 charges at 3 corners: Q, 3Q and -2Q. The question is what is the potential at the fourth corner. We just need to remember that the potential V® due to a single charge is:
$V(r)=\frac{q}{4\pi\epsilon_0 r}$.

$V_t=\sum_i V_i=\frac{Q}{4\pi\epsilon_0 \sqrt{2} l}+\frac{3Q}{4\pi\epsilon_0 l}-\frac{2Q}{4\pi\epsilon_0 l}$

### 23.36

We need to integrate the potential due to a differential of charge $d$q. $V=\int{\frac{\partial q}{4\pi\epsilon_0 r}}$.
$r=l/\pi$
$V=\frac{\pi}{4\pi\epsilon_0 l}\int\partial q=\frac{Q}{4\epsilon_0 l}$.

### 23.61

$\frac{v_e}{v_p}=\sqrt{\frac{m_p}{m_e}}$

### 23.84 