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__Electrostatics (chapter 15)__

__Formulas:__

Charge on an electron = e = -1.602 x 10^{-19}C
(proton is same, only positive)

Charge = q = ne (where n= # of electrons. e = charge on electron) (Units = Coulombs = C)

Coulombs’ Law = F_{e} = kq_{1}q_{2}/r^{2}
where k = 9 x 10^{9 }Nm^{2}/C^{2}

Electric field = E = F_{onqo}/q_{o} = kq/r^{2}
= 4pkQ/A
= Q/Î_{o}A (where Î_{o} = 8.85 x 10^{-12}
C^{2}/Nm^{2} = -DV/Dx_{max} (where Dx_{max} = maximum
distance between two plates)

Electric field from a point charge (q) = E = kq/r^{2}

__Things to remember:__

* *

1. Newton’s laws still apply. Recall that F = ma

2. Point charges (q_{o})
are usually assumed to be positive unless otherwise stated.

3. Moving towards an area of positive charge increases potential. Positive test charges tend to move towards areas of LOWER potential (they move towards negative charges/areas)

__Capacitors/Dielectrics (chapter 16)__

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__Formulas:__

Electric field between two closely spaced parallel plates =
E = 4pkQ/A
= F/q_{o}

Electric potential difference = voltage = DV
(units: Volts = J/C = Nm/C = Kgm^{2}/Cs^{2})

DV
= DU/q_{o}
= W/q_{o} (where W = work = Fd) = Ed = kq/rb – kq/ra = Q/C = q_{o}Ed/q_{o}

Voltage = V = kq/r

Potential energy = DU = work = F cosq d =
Eq_{o}d (units: Joules = N/C =
Nm)

Potential energy of a capacitor = U_{c} = 1/2 CV^{2}
= 1/2QV = 1/2Q^{2}/C = work = Fcosqd = Eq_{o}d = area
under a voltage vs. charge(q) graph

Work = Fcosqd = DU = DKE = -qDV = eDV (where e = -1.602 x 10^{-19}C)

(units: Joules = N/C = Nm)

Capacitance = C = Q/V = Q/Ed = ÎoA/d = Qr/kq = 1/slope of a voltage vs. charge (q) graph (Units: Farad = Coulomb/Volt = C/J/C)

__Things to remember:__

1. Voltage is basically the work (Fcosqd) done to moving a point charge (qo) between two points.

2. Voltage increases (becomes more +) when you move closer to positive objects or away from negative ones.

3. Voltage itself is not usually an interesting value. We are usually concerned with DV. Why? Because you define an electric field with two points. Think of an electric field like a gravitational field. When you stand on the ground, your potential is zero. When you stand on a chair, your potential is = mgh where h is defined by TWO points (1) the ground and (2) the height of the chair above the ground.

Similarly, an
electric potential difference (Voltage) is defined as the distance (h) between
two electrically charged plates (they usually use “d” for this separation
distance of charges or plates and leave the “h” for height in a gravitational
field). The main difference between electricity
and gravity is that in a gravitational field, all objects are attracted in __one
direction__. In an electric field,
depending upon the charge of the object, it can be attracted one way or the
other.

4. You can think of DV as being independent of path taken from point A to point B because it is the net distance difference between point A and point B which determine the voltage potential. Think of electric potential as a series of infinite concentric rings around a point charge. These are areas of equipotential (p. 514-520)

5. So what is the difference between DU and DV anyway?

DU = potential energy for ALL particles in a field (be it gravitational or electrical)

DU_{gravity}
= mgh

DU_{electrical
}= q_{o}Ed (Notice how
these symbols are analogous to mgh)

DV = potential difference for ONE SINGLE particle in a field (be it gravity or electrical)

DVgravity (if there were such a thing) = gh

DVelectrical
= q_{o}Ed/q_{o} = Ed

When talking about gravity, we are usually interested in the potential energy (or work) done to an entire object of mass m as it is lifted in the field (mgh). In gravity, we are usually NOT concerned with the strength of the field at some given height (in other words, we don’t care about gh).

When talking about electricity, we are usually interested in the force that would be applied to every single little bit of electricity in the field (Ed). It is like subtracting out the mass because DV is independent of the mass/charge of the particle in the field. Unlike gravity, we ARE concerned about the strength of the field at a given point (here we DO care about gh as it were).

6. Series Capacitors
(1/C_{s} = 1/C_{1} + 1/C_{2} + 1/C_{3}…… where
C_{s}<C_{i}) C_{i
}= individual capacitor in the series)

Q: Charge is the same on all capacitors = CiVi

V: Sum of each V_{i
}= V_{battery}

C: C = Q/Vi where V_{i}
= V_{battery}

7. Parallel
Capacitors (Cp = C_{1} + C_{2} + C_{3}….. where C_{p}>C_{i})

Q: charge = V_{battery}(C_{i})

V: Voltage is the
same for all so V_{i} = V_{battery}

C: C = Q/V_{i }where
V_{i} = V_{battery}

8. As distance
between two charged plates increases:
Potential energy (U_{c}) decreases and Capacitance (C)
increases. In otherwords, the better a
capacitor is at holding a charge, the worse the potential energy it can
provide. That makes sense.

__Electric Circuits (chapter 17)__

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__Formulas:__

Emf = Voltage potential between two terminals of a battery that are NOT connected to anything yet. Emf – V = IR

Current = I = q/t (units: C/sec = Amp)

Resistance = R = V/I (units: volt/amp = ohm = W)

Voltage = V = IR (ohm’s law)

Power = P = IV = V^{2}/R = I^{2}/R

Resistivity of wires/resistors = r = RA/L

__Things to remember:__

1. Series
resistors: R_{s }= R_{1}
+ R_{2} + R_{3}…. Where R_{s}>R_{i}

V_{battery} = IR_{s} (where V_{battery}
is constant)

So, as R increases, total Resistance increases and current (I) decreases.

2. Parallel
resistors: 1/R_{p} = 1/R_{1}
+ 1/R_{2} + 1/R_{3} ….. Where R_{p}<R_{i}

V_{battery} = IRs (where V_{battery} is
constant)

So, as the number of resistors increases, the total resistance goes DOWN and current (I) goes up.

3. If you need help remembering how to find “equivalent” resistance, look at fig 18.6 on p. 574.

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__Formulas:__

Force on a moving **charged particle** = F_{q} =
qvBsinq
(where v = velocity and q = angle between velocity and the B field.)

Force on a long **straight wire = **F_{w} =
ILBsinq
(where q
is the angle between the wire and the B field and I is the current and L is the
length of the wire)

Torque on a single current carrying loop = t =
IABsinq
(where A = area of loop and q is the angle between a **NORMAL** line to the plane
of the loop and the B field)

Magnetic flux = F = BAcosq (where q = the angle between a NORMAL to the plane of the loop and the B field)

Induced Electromotive Force on a wire by a magnet = EMF = -N (DF/Dt) (where DF = change in flux)

B field on a long wire = m_{o}I/2pd

B field in the center of a loop = m_{o}I/2r

B field in the center of a solenoid = m_{o}NI/L
= m_{o}nI (where n = # of turns/length)

__Things to remember:__

**Faraday’s Law**:
You can induce an EMF by changing the quantity of B field lines passing
through a loop of wire. F =
BAcosq and EMF = -N (DF/Dt)

**Lenz’ Law:** An
induced EMF in a metal loop (or coil) gives rise to a current who’s direction
is OPPOSITE such that its induced magnetic field is OPPOSITE to that which
produced it. This makes sense because
of the law of conservation of energy.
Let’s say, for the sake of argument, that an induced magnetic field is
in the same direction as that which produced it. That means that the net magnetic field would have to be additive
– and if you have learned one thing in physics, I hope that it is: You can’t
get something for nothing (First Law of Thermodynamics!)

**Right hand rules:**
Read them on **p. 625** under **IMPORTANT EQUATIONS**. Be sure to note when your thumb is the
direction of force and when your thumb is the direction of current. __There is only one of the three cases
where your thumb is the current.__

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