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Goal: To understand what electric force is and how to calculate it.

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### Goal: To understand what electric force is and how to calculate it.

Objectives:

Understanding how to translate electric field to force

Understand how to calculate Electric forces

Knowing what Electric Field lines are and how to use them

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 / r2
- 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 * qon
- Where qon is the charge the force is being applied to and E is the electric field that charge qon 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 / r2
- So, F = -qon * qby * k / r2
- (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 / r3 (x hat)
- Fy = -k qon * qby * y / r3 (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.

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