Transcript Goal: To understand what electric force is and how to

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 / 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
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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
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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
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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.