Download 4000 N/C

Think
• We calculated the force of gravity for 2
electrons to be much much smaller than
the electrostatic force between these two
electrons. Would it be possible for a
charged object to have an equal force due
to gravity (downwards) and an
electrostatic force pushing upwards?
Explain !
Test Tomorrow !
• Wondering why you did not get in the 90’s
on the last quiz?
• Think back to how seriously you did the
homework.
• Make sure you review/do the homework
assignments before the test. They should
all be open. Study- make sure they make
sense.
Today
• Review by looking at a topic that combines
several ideas. Make sure you are asking
questions if it does not make sense !!
• HW – Castle Learning – Electrostatic
Review (and ALL unfinished homeworks)
Millikan’s Idea
• Robert Millikan was a scientist who studied
electricity.
• During the early 1900’s, scientists thought
of electricity as a fluid.
• They had the concept of CHARGE, but not
of ELECTRONS!
• Millikan set about to find out if there were
‘bits’ of charge
Things Millikan Knew
• Acceleration due to gravity
• An object with a net force of 0 moves with a
constant velocity
• Droplets can be charged using x-ray’s
• “Stokes’ Theorem” – mass of droplets
• How to find the electric field between two plates
Millikan’s Apparatus
Millikan sprayed oil droplets into a chamber with
charged plates at the top and bottom
The droplets come out with a CHARGE on them
They areThey
attracted
to one plate or the other
are attracted to one plate or the other
depending on their charge!
Millikan’s Apparatus
Millikan noticed that when he adjusted the electric field,
some of the droplets would not move UP or DOWN.
WHY???
Fg must be equal to Fe
forces
on an
object
WhatWhat
do we
knowact
about
these
twoinforces?
gravitational and electric fields?
Equations
Millikan looked at this using
the equations:
Fe  Eq
Fg  mg
He noted that droplets
only stopped moving
when the electric field
was set to
certain strengths.
Eq  mg
mg
q
E
What does this suggest?
Electrical “Bits”
• Millikan did many observations, collecting
thousands of data samples for charge on oil
droplets.
• This experiment has been repeated by hundreds
of physicists (and university physics students)
with the same results.
• Charge ONLY occurs in
multiples of 1.6 x 10-19C
• A +5.0 coulomb charge is located in a uniform
electric field of intensity 4000 newtons per coulomb.
Calculate the force on the object.
E = 4000 N/C
+ 5.0 C
F = Eq
F = (4000 N/C)(+5.0 C) = 20,000 N
• Which of the following is a correct drawing of
the field lines near a negatively charged
object?
a
b
c
d
b.
Field lines pointing INWARD
- - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + +
• A proton is placed between two equal, but
oppositely charged plates. What will happen
to the proton?
+
The proton will ACCELERATE toward the negative plate!
The electric field between the plates provides a constant,
unbalanced FORCE on the proton
• At which point is the electric field intensity
ZERO?
A
+ 2.0 C
B
+ 2.0 C
At point B.
The midpoint between two like charges experiences no ‘net
field’ – the fields of the like charges cancel out in that area.
• In which direction will the electric field
between the plates point?
+ + + + + + + + + + + + + +
- - - - - - - - - - - - - - - - - - Field lines always point from the positive plate to the
negative plate.
• An object with a charge of -0.45 coulombs
experiences a force of 45 newtons when placed
in an electric field. What is the electric field
intensity at this point?
Fe = 45 N
-0.45 C
E=F/q
E = 45 N / -0.45 C = -100 N/C
• Rank the points in order from greatest electric field
intensity to least.
C
A
-
D
B
D, A, C, B
Field strength gets weaker as we get farther away from the
charge.
• Which electron will experience the greatest
acceleration?
- - - - - - - - - - - - - - - - - - A
B
C
+ + + + + + + + + + + + + +
All three electrons experience the same electric field
intensity – this means that they encounter the same force,
and so the same acceleration!
• What is the direction of the resultant electric
field vector at point A?
A
+ 2.0 C
+ 2.0 C
Toward the top of the screen. A test charge placed at point
A will not move left or right because it is equally repulsed
by both charges in that direction. However, BOTH
charges will push it UPWARD.
• An electron passes through the electric field
between two parallel plates. In which
direction will it be deflected?
- - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + +
The electron will deflect toward the positive plate.