Transcript Electrostatics Part I

Part 3
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The electric field can push AND pull charges
 Because there are two types of charges (+ and -)
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The gravitational field can only pull
 Only positive masses
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Charges cause an electrical field.
Each charge creates its own electric field in the
surrounding space.
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Electric Field is a vector quantity
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 Has magnitude and direction
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Use Electric Field lines(lines of force) to show the
Electric Field
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Electric field direction from charge that creates the
field:
 Away from positive charge
 Toward negative charges
The lines begin on positive charges and end on
negative charges.
 Draw lines perpendicular to charge
 Lines never cross
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Electric field strength:
 The closer to the charge, the stronger the field
 The closer the field lines, the stronger the field
 The farther away, the weaker the field
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Electric field strength:
 The number of lines are proportional to strength
 The more lines the stronger the field
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Electric field strength: (quantitative)
E=
2
kq/d
▪ Depends on Charge and distance
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E = Electric field strength (Newton/Coulomb, N/C)
k = 9.0 x 109 Nm2/C
q= charge causing the field (Coulomb, C)
d or r = distance (meters, m)
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A fly accumulates 3.0 x 10-10 C of positive charge
as it flies through the air. What is the magnitude
and direction of the electric field at a location 2.0
cm away from the fly?
q = 3.0 x 10-10C
d or r = 2.0 cm = 0.020
_____ m
E=?
equation E = kq/ r2
E = (9.0 x 109)(3.0x10-10)
(0.020)2
E = 6750 N/C
Electric Field = force exerted on positive charge
If +q = force in same direction (pull) of the field
If -q = force in opposite direction (push) of the field
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The strength of the electric field…
E = F/q
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E = strength of the field (Newton/Coulomb, N/C)
F = Force exerted on charge in field (Newton, N)
q = charge (Coulomb, C) that sits in the field;
experiences the force
A charge of -2.0 x10-6C is placed in a uniform
electric field of strength 5000 N/C that points
downward. What is the magnitude and direction of
the force experienced by this charge?
A charge of -2.0 x10-6C is placed in a uniform electric field
of strength 5000 N/C that points downward. What is
the magnitude and direction of the force experienced
by this charge?
E=F/q
5000 N/C= F / (-2.0 x 10-6C)
(-2x10-6 )(5000)=F
-0.01N=F magnitude of the force
direction? - 2 charge, the force on the electron is opposite the field!