What is current?

It is a flow….but what is flowing?

Electrons!

How do we measure the flow rate?

How can we measure how many electrons are flowing?

Measure charge!

What is the unit for charge?

Coulomb (C)

But if we are examining a rate, what else do we need to take into account?

Time!

Therefore, we are examining the quantity of charge per unit time or C/s.

And….a C/s is called an Ampere or Amp (A) for short.

+Q

q

q

q

Q is doing something to the “space” around it…the force on q depends on where you put it in this “space”.

If we know what a point in “space” is like, we can describe the force on some charge q in that space.

As q is moved further away, what happens to the force?

Now, let’s try to understand voltage……

kqQ

r2F =

Coulomb’s Law

kqQ

r2F = kQ

r2 = q[ ]

describes the “space” around q

We call this space the Electric Field, E where E = kQ

r2

Therefore, F = qE or E = F/qwhere q is defined to be a positive test charge.

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+

+

+

+

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d

qF

Uniform Electric Field (with fringing around edges)

Field Lines 1. Leave +, Enter – 2. Perpendicular to Conductor

Potential Energy = 0

If the charge is free to move, what will happen?How much work is done by the force acting on the charge?

Work done = Fd = qEd = ½ mv2

Work done Change in energy

If q was placed at the right hand side (RHS), what would happen?

What if you want it to reach the left hand side (LHS)? What would you need to do?

WORK! (like lifting up an object)

How much work do you need to do?

Work = Fd = qEd = ΔPE

What if you placed q’ at the RHS where q’ = 2q? How much work do you need to do?

Work = 2qEd

For a given charge, how much energy does it take (or how much energy do you get out) when moving from one plate to another?

0ΔPE = -q Ed (valid only for uniform electric field)

Potential difference between the initial and final point is defined as the change in potential energy of a charge divided by the charge.

Use V to represent potential difference.

ΔV = Vfinal – V initial = ΔPE/q ΔPE = qΔV

Look at the units of ΔV…

[ΔV ] = [ΔPE]/[q]

[ΔV ] = J / C ….. We refer to this as a Volt

In other words, 1 J of work must be done to move a 1 C charge between two points that are at a potential difference of 1 V.

In the process, the 1C charge gains (or loses) 1 J of energy as it moves through the potential difference.