1. Electric Potential

2. Potential Difference
Suppose we have an electric field in space created by stationary charges somewhere
So we have an electric field looking like this (draw it) and it can vary (non uniform)
Let’s consider 2 points, A & B and we want to define the potential difference between points A & B
Now let’s say we have a point charge, q0 that moves from A to B along some path.
As it moves from A to B the electric field exerts a force on it.
E-field E exerts a force given by q0. E = F
This is very similar to a mass moving in a gravitational field.
We define the potential energy in terms of gravity, because the force of gravity is conservative
This is the same for the electric force. (the work done is independent of the path)
(Draw additional line on pic)
So we can define the electric potential energy (U) is defined as
ΔU = -int (F.ds) where ds is a tiny segment along the line. Where ds = a the line integral.
We take the path and divide it into small chunks of ds1, ds2, …,dsN etc.
We take the force at the midpoint of dsi as it doesn’t change
We evaluate Sum Fi.dsi all the way up to as N  infinity.
If the force is conservative, the line integral is independent of path.
We can take q0 out of the equation.
Now we want an equation that is independent of the charge q0
So we define this new term called ΔV called the electric potential.
ΔV = int E.ds
If the electric field is 0 then they are at the same potential.
If E and ds are perpendicular then A and B are at the same potential.
The work done by the net force gives rise to the change in kinetic energy.
To keep the velocity the same the net force must be 0.
Work done by external agent W = ∫F.ds = -q0∫E.ds
Lift it very slowly so at height h the speed is 0.
Where did this work go? It goes into potential energy.
So if I move it from A to B and the KE remains the same, then the work is add goes to increasing the PE.

3. Potential Difference in a Uniform E Field
Draw field lines with points A and B.
For simplicity we take A and B horizontal like the lines.
Then what is ΔV = Vb-Va = -∫E.ds
But E is constant and ds is the same direction of E so we lose the vector part. (cos theta)
V = -Ed (mention volts / metre) as we did before.
Think of the flow of water, as we go from A to B the field is to the right, so we move with the flow. We move downward. So the potential is decreased. Such as a boat down the river. So A is a higher potential than B.
Va-Vb is positve. (Ed)
Vb-Vba is negative. (-Ed)
But what happens if A and B are not parallel, lets say I want to go to point C instead.
What is Va-Vc = -∫E.ds from A to C
Again E is constant. But this time E and ds are no longer parallel, but the angle between them is constant.
-∫E.ds = -∫E cos(theta) ds again we can take them outside the integral.
Again we get the length of the line from the integral.
But this is simple –E.d
Notice if B and C are at the same horizontal distance. (BC perpendicular to A) Then ΔV from A to B and A to C is the same.
In fact if I take a plane then all the point on the plane will have the same electric potential. The same as gravity, an item raised any height will have the same potential energy.
This is called an equipotential. All points on the surface have the same electric potential.

4. Parallel Plates
Let us do an example. Let’s say we have parallel plates like in a capacitor.
If I make the approximation that the plates are infinite we know the electric field inside is.
Recall is we have a plate, the electric field is perpendicular and the E field is σ/2ε0 where sigma = surface charge density. This expression is exact if the sheet is infinite. σ = Q/A
Lets put another plate that is negative. The field is the same σ/2ε0
Inside the fields add up to σ/ε0 and outside the fields cancel.
So inside the plates we have a field given by σ/ε0. This approximation is fine aslong as d << A
This is not quite true nearly the edges (but we ignore that)
Now if I pick 2 points A and B.
What is the potential difference ΔV = -Eh where h is the change in height.
Now if we put a point charge in there. If it moves from A to B it’s potential energy changes.
ΔU = Ub-Ua = qΔV = -qEh
It’s negative as B is a lower potential than at A. As we are moving with the flow. Similar to going to down in height in gravity.
If we take a charge q that we release from rest from A. so it’s speed at A is 0.
Because of the E field there is a force on q acting down. The particle will accelerate down. A = qE/m
What is the velocity at point B?
This problem is idential to that of gravity and an item falling.
We can use conservation of energy.
KA + UA = KB + UB
0+UA=1/2mvB2 + UB
1/2mvB2 = qEh
Rearrange to find vB

5. Motion of a Charge Particle in a Uniform E Field
Lets say we shoot it in at the side this time.
This is the same as projectile motion
We get a parabola.
The force qE is down which is constant. (a=qE/m)
There is no horizontal acceleration.
We can find the velocity at any point that it has descended a certain height.
We can use conservation of energy.
1/2mv02 + Ui = 1/2mvf2 + Uf
1/2m(vf2-v02) = Ui-Uf = qEh
We can find Vy = root (v2 – vx2) = root (v2-v02)

6. Homework Do the sheet please  Due in 19th February
Read p676 to end of chapter