## Important Points from Lectures

1)**Choosing Gaussian Surface:**When you are working on a problem where you want to use Gauss' Law to find a field, there are a few issues you want to be aware of:

- a. The law only tells you the surface integral of the field (the flux) - if the field is varying on your surface in a way you don't know before hand, this can become very intractable. It is best to set up your surfaces so that you have the same field everywhere if possible.

b. Remember also that the integral only counts the perpendicular component of the field, so you need to have the field that you're interested in perpendicular to the surface.

c. But, we can use this to our advantage: if your surface needs to include an area where the field is either not the value of interest, or not constant or both (eg: long line of charge), if your surface passes through this space in such a way that the field is parallel to it, then you can safely ignore these fields!

**Conductors:**Field at surface must be perpendicular to surface, field inside must be zero. (Even for hollow ones: eg Faraday Cage experiment).

3) **Field Lines:** Another way to draw in fields. They show direction of electric field directly and tell you the strength of the field with their density. Field Lines cannot cross (but they can run from one charge to another, or meet head on at a point where field is zero).

4) **Electric Potential and Potential Energy:** Work done by field on a charge as it moves is equal to the drop in electrical potential energy between the starting and finishing points (Electric field is conservative force field). Here V is used to denote the potential energy of a specific charge in the presence of other charges and φ(r) is the potential at a point in space due to an arrangement of charges.

- a. Two point charges: V(r) = kq

_{1}q

_{2}/r

b. φ(r)=V(r)/q_{2}

c. Superposition still works: V=Σ_{i}kq1qi/ri1 is energy of charge one Ã¢â‚¬â€œ our test charge in most cases, (and there is a similar sum to calculate the "self energy" of the fixed collection of charges).

- a. Coulomb's Law: F=kQ

_{1}Q

_{2}

**ρ**/r

^{2}b. Electric Field of a point charge: E=kQ

**ρ**/r

^{2}c. Electrical Potential Energy (scalar): EPE=kQ

_{1}Q

_{2}/r d. Electrical Potential (also scalar): φkQ/r

**2ρ**is a unit vector pointin in the direction of

**r**.

6) **Potential Energy:** Given a fixed charge distribution, how much work do I have to do to bring a single charge in from infinity to a particular point? Alternately, how much kinetic energy will my test charge have once it reaches a very large distance away, assuming it was released from rest?

7) **Electron Volts:** 1eV is the energy an electron would have after it is accelerated through one volt of electrical potential.

8) **Equipotential surfaces:** imaginary surfaces where φ has a fixed value - electric field is perpendicular to these surfaces, just like a conductor.

9)**Electric Dipole:** two slightly separated, equal but opposite charges:

- a. φ=k|p|cos(ϑ)/r

^{2}

b. p=qd - Dipole moment

**Capacitance:**Q=CV, C=ε

_{0}A/d (for parallel plates), U=1/2 QV =1/2 CV

^{2}=1/2 Q

^{2}/C

## Example Problem 1

How strong is the electric field between two parallel plates 5.2mm apart if the potential different between them is 110V?Definition of potential energy: work done moving charge is equal to the change in potential. If a one coulomb charge moves from one plate to the other, the field does work, W, equal to 110J

- W=110V*1C

W=110J

- W=Fd

W=E*1C*5.2x10^{-3}m

E=110J/(1C*5.2x10^{-3}m)

- E=2.1x104V/m

## Example Problem 2

An electron starts from rest 72.5 cm from a fixed point charge with Q=-0.125μC. How fast will the electron be moving once it reaches a very large distance?Initially the system (consisting of the electron and the fixed charge) have only potential energy:

- PE = kQ

_{1}Q

_{2}/r

PE = 2483x10^{-19}J

- KE = 1/2 m

_{1}v

_{1}

^{2}+1/2 m

_{2}v

_{2}

^{2}

KE = 1/2 m_{2}v_{2}^{2}

- 1/2 m

_{1}v

_{1}

^{2}=PE

v=23x10^{6}m/s

## Example Problems 3 and 4

How strong is the electric field between the plates of a 0.8μF air-gap capacitor if they are 2mm apart and each has a charge of magnitude 72μÂC? Q=CVV=Q/C=90V

E=V/d = 45000V/m

A capacitor C1 carries a charge Q_{0} initially. Then it is connected in a circuit with a second, uncharged capacitor: C2. What charges do each of the capacitors carry now? What is the potential difference across them?

The instant that the capacitors are connected, charge flows to even out the voltage (the wires are conductors, therefore there can be no potential drop across them in this static case).

- |V1|=|V2|

Q_{1}/C1=Q_{2}/C2

Q_{1}= (Q_{0}-Q_{1})/C2

(C2-C1)Q_{1}=C1 Q_{0}

Q_{1}=C1Q_{0}/(C2-C1)

- Q

_{2}=C2Q

_{0}/(C2-C1)

V= Q_{0}/(C2-C1)

## Example Problem 5

Near the surface of the earth, there is an electric field of E=150V/m pointing downwards. Two ball with opposite charges (|q|=550μÂC) that are otherwise identical (m=.54kg) are dropped from a height of h=2m. With what speeds do they both hit the ground?Conservation of energy:

- mgh+ E*h*q=1/2mv

^{2}

v=(2gh+2Ehq/m)^{1/2}

_{+}=6.37m/s

for the negative ball" v_{-}=6.27m/s

as a comparison an uncharged ball will hit with v_{0}=(2gh)^{1/2}=6.32m/s

A positively charged ball will experience an extra downwards force in this field and thus hit faster that the neutral ball, which will in turn hit faster that the negatively charged ball which will be partially "levitated" by the electric field.

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