Goal: To understand what electric force is and how to

calculate it.

Objectives:1) Understanding how to translate electric

field to force2) Understand how to calculate Electric

forces3) Knowing what Electric Field lines are and

how to use them4) 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.