1. Forces From Continuous Charges
If you have a spread out charge, it is tempting to start by calculating the total charge
Generally not the way to go
The charge of the line is easy to find, Q = L
But the distance and direction is hard to find
To deal with this problem, you have to divide it up into little segments of length dl
Then calculate the charge dQ =  dl for each little piece
Find the separation r for each little piece
Add them up – integrate
For a 2D object, it becomes a double integral
For a 3D object, it becomes a triple integral q r  dl

2. The units for electric field are N/C
The Electric Field
Suppose we have some distribution of charges
We are about to put a small charge q0 at a point r
What will be the force on the charge at r?
Every term in the force is proportional to q0
The answer will be proportional to q0
Call the proportionality constant E, the electric field q0 r
The units for electric field are N/C
It is assumed that the test charge q0 is small enough that the other charges don’t move in response
The electric field E is a function of r, the position
It is a vector field, it has a direction in space everywhere
The electric field is assumed to exist even if there is no test charge q0 present

3. Electric Field From a Point Chargeq q0
From a single point charge, the electric field is easy to find
It points away from positive charges
It points towards negative charges - +

4. Electric Field from Two Charges
Electric field is a vector
We must add the vector components of the contributions of multiple charges + + - +

5. Electric Field Lines + -
Electric field lines are a good way to visualize how electric fields work
They are continuous oriented lines showing the direction of the electric field + -
They never cross
Where they are close together, the field is strong
The bigger the charge, the more field lines come out
They start on positive charges and end on negative charges (or infinity)

6. Electric Fields From Continuous ChargesP
If you have a spread out charge, we can add up the contribution to the electric field from each part
To deal with this problem, you have to divide it up into little segments of length dl
Then calculate the charge dQ =  dl for each little piece
Find the separation r and the direction r-hat for each little piece
Add them up – integrate
For a 2D object, it becomes a double integral
For a 3D object, it becomes a triple integral r  dl

7. Sample Problem Divide the charge into little segments dl y
Because it is on the y-axis, dl = dy
The vector r points from the source of the electric field to the point of measurement
It’s magnitude is r = y
It’s direction is the minus-y direction
Substitute into the integral
Limits of integral are y= a and y = 
Pull constants out of the integral
Look up the integral
Substitute limits
y=a x y
What is the electric field at the origin for a line of charge on the y-axis with linear charge density (y) = Qy2/(y2+a2) stretching from y = a to y = ? dy
(y) r P