Goal: To understand what electric force is and how to calculate it. Objectives: 1)Understanding how to translate electric field to force 2)Understand how to calculate Electric forces 3)Knowing what Electric Field lines are and how to use them 4)Understanding motions of a charged particle in a constant electric field.

Yesterday: We learned that the Electric field is a topography of electric charges around you. At any point the electric field is just a sum of the topography from each charge. For each charge E = -qk / r 2 How would this translate to a force?

Ball downhill If you have a gravitational topography a ball will want to roll downhill. That is it will roll from a high elevation to a low one or a high field to a low one. The same is true of electric fields. A positive charge will want to move to a lower electric field. A negative charge will do the opposite and will want to move up to a higher valued electric field (moving uphill).

Now for the math The force on a charge is: F = E * q on Where q on is the charge the force is being applied to and E is the electric field that charge q on is located at. Much like for gravity that F = m * g on the surface of the earth.

If we add in E If we have 2 charges called qon and qby then the force is: F = qon * E, but E = -qby k / r 2 So, F = -qon * qby * k / r 2 (k is the same constant we had before) And if there are more than 2 charges, each charge will have a force on qon. The net force will add up just like you add them up for E.

Using the vectors The vector way to find the force: Fx = -k qon * qby * x / r 3 (x hat) Fy = -k qon * qby * y / r 3 (y hat) Sanity check: like charges repel and opposites attract. The sign and direction should reflect that.

2 dimensions Just like yesterday in 2 dimensions you have to take the dimensions into account. We will start off with a straightforward 3 charge problem. q2 = 5 C and is at y = 3, X = 0 q3 = 9 C and is located at y = 0, x = 6 What is the total force on q1 if it is at the origin and has charge of 3 C?

Now we take the next step Now a little bit harder. q2 = 3 C is at y = -2, x=0 q3 = -5 C and is at x = 3, y = -4 q1 = -2 C and is at the origin What is the vector form of the force and what is the magnitude of the force on q1?

Field lines Another way to look at this is by looking at field lines. Field lines point downhill – the direction a positive charge will flow. While these lines will tend to move towards – charges and away from + charges, that is not always the case if you have many charges. (draw on board)

Motions of a charge in a uniform electric field Imagine you have an entire room where at any point in that room the electric field is about the same. If you put a charge into that room then what will the charge do? A) do nothing – no movement B) move around in a circle C) move around the room in random way D) accelerate in some direction at a constant rate E) accelerate in some direction in an ever increasing rate

Motions of a charge in a uniform electric field Imagine you have an entire room where at any point in that room the electric field is about the same. If you put a charge into that room then what will the charge do? A) do nothing – no movement B) move around in a circle C) move around the room in random way D) accelerate in some direction at a constant rate E) accelerate in some direction in an ever increasing rate Since F = q * E that already tells you the force will be a constant because q and E are constant here. Also, ALWAYS remember that F = ma… So, F = q * E = ma So a = q * E / m for a uniform electric field! Thus the acceleration is constant and the direction will be determined by the charge and the direction of the electric field.

Conclusion F = q * E Electric Field lines point downhill. If E is uniform then F and a are constants! Once again the hardest part is doing the geometry.