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Electric Field
Contents:
•Basic Concept
•Calculating electric Field example
•Whiteboards
•Point Charges
•Examples
•Whiteboards
•Electric Fields in Charge Arrays
Gravitational Field
Electrical Field
Force on mass
g = Field strength
g = 9.81 N/kg (on earth)
F = mg
Force on charge
E = Field strength
E is in N/C
F = qE (E = F/q)
Direction: Uh… down
Direction: F on + Charge
Force:
F = Gm1m2
r2
Field:
g = F = GM
m
r2
Force:
F = kq1q2
r2
Field:
E = F = kq
q
r2
G = 6.67x10-11 Nm2kg-2
k = 8.99x109 Nm2C-2
Handout/Get out your data packet!!
Electric Field
Example 1 - A +125 C charge experiences a force to the
right of 0.0175 N. What is the Electric field, and its
direction?
E = F/q = 0.0175 N/125x10-6 C = 140 N/C to the right
Direction:
+q
Force This Way
E
Force This Way
-q
Which way is the electric field? (wwpcd?)
+
+
+
+
+
-
Which way is the electric field?
-
+
+
+
+
+
Which way is the electric field?
+
Which way is the electric field?
+
-
Electric Field
Example 2 - An electron travels through a region where
there is a downward electric field of 325 N/C. What
force in what direction acts on the electron, and what is
its acceleration?
F = Eq = (325 N/C down)(-1.602x10-19 C) = -5.21x10-17 N down
F = ma, a = F/m = (5.21x10-17 N)/(9.11x10-31kg) = 5.72x1013 m/s/s
Whiteboards:
Uniform field
1|2|3|4
Ishunta Dunnit notices that a charge of
-125 C experiences a force of 0.15 N to
the right. What is the electric field and
its direction?
E = F/q = (.15 N)/(-125x10-6 N) = -1200 N/C right
or 1200 N/C left
1200 N/C left
Doan Botherme places a +12 mC charge
into an upward 160 N/C electric field.
What force in what direction does it
experience?
E = F/q, q = 12x10-3 C, E = 160 N/C up
F = 1.92 = 1.9 N up
1.9 N up
Alfred O. Dadark is on a planet where a
mass of 0.12 kg experiences a downward
force of 7.80 N. What is the gravitational
field on the surface of this planet?
g = F/m, m = 1.12 kg, E = 7.80 N down
g = (7.80 N down)/(0.12 kg) = 65 N/kg down
65 N/kg down
Telly Vishun places an unknown charge
into a known upward electric field of 612
N/C, and the charge experiences a
downward force of 0.851 N. What is the
charge?
E = F/q, E = 612 N/C up, F = .851 N down
q = -.0013905 C = -1.39 mC
-1.39 mC
Sal F. Hone levitates a 0.00125 kg ball
with an upward electric field of 590 N/C.
What is the charge on the ball?
(Hint gravity = electrical force)
Eq = mg
E = F/q, F = Eq, F = mg, m = .00125 kg, g = 9.80
N/kg, E = 590 N/C
q = 2.07627E-05 = +20.8 C
+20.8 C
Electric fields around point charges
Direction of E field is:
Away from positive
Toward negative
Gravitational fields around mass
Direction of g field is:
m
Toward mass
Fields around points of mass or charge
Example: What is the electric field 2.0 m
above of a -21 μC charge?
E = kq/r2 =
(8.99E9)(-21E-6)/(22) =
-47,197.5 N/C (away) or
47,000 N/C toward, or downward
a, b, c are points,
Q is a charge
Whiteboards:
Field relative to points
1|2|3
Vera Similitude measures the electric
field 13.5 m to the right of a -1.45 C
charge. What electric field in what
direction?
E for a point charge:
E = kq
r2
k = 8.99x109 Nm2C-2, q = -1.45x10-6 C, r = 13.5 m
E = -71.5 N/C away from the charge, or to the left (toward
the negative charge)
71.5 N/C to the left
Vesta Buhl measures an electric field of
2,120 N/C upward, 67 cm above a charge
of unknown value. What is the charge?
E for a point charge:
E = kq
r2
k = 8.99x109 Nm2C-2, E = 2,120 N/C, r = .67 m
q = 1.06x10-7 C = +.11 C. It is a positive charge as
the E-field is away from it
+.11 C
Amelia Rate measures a gravitational
field of 3.4 N/kg. What distance is she
from the center of the earth? (Me = 5.98
x 1024 kg. )
g for a point mass:
g = Gm
r2
G = 6.67x10-11 Nm2kg-2, g = 3.4 N/kg, m = 5.98x1024 kg
r = 10831137.03 m = 10.8 x 106 m (re = 6.38 x 106 m)
1.1 x 107 m
Tara Bull measures an electric field of
10. N/C what distance from an electron?
E for a point charge:
E = kq
r2
k = 8.99x109 Nm2C-2, E = 10 N/C, q = 1.602x10-19 C
r = 1.20008x10-5 m = 12 m
12 m
+12.5 C
x
A
72.1 cm
-187 C
C
11.2 cm
What is the electric field at the x?
(Magnitude and direction)
EA = kqA = 8.958E6 N/C to the left (away from +)
r2
EC = kqC = 2.423E6 N/C to the right (toward -)
r2
= - 8.958E6 N/C + 2.423E6 N/C = -6.54E6 N/C (Left)
6.54E6 N/C to the left
Each grid is a meter. If charge A is -1.50 μC, and charge B is +1.70 μC,
calculate the electric field at the origin: (1649 N/C up and right 69.4o with the x axis)
y
-
A
x
B
+
Electric Fields in Non Linear Arrays
A
+1.5 C
75 cm
Q1
Find the electric Field at point A:
+3.1 C
190 cm
1. Calculate the individual E-Fields
2. Figure out direction (angle)
3. Add E-Fields (as vectors)
Q2
Electric Fields in Non Linear Arrays
E1
E2
A
+1.5 C
75 cm
Q1
1. Calculate the individual E-Fields
2. Figure out direction (angle)
3. Add E-Fields (as vectors)
+3.1 C
190 cm
E1 = kq1 = k(1.5 C) = 11,987 N/C
r2
(.752 + .752)
E2 = kq2 = k(3.1 C) = 6,679.2 N/C
r2 (1.92 + .752)
Q2
Electric Fields in Non Linear Arrays
E2 = 6,679.2 N/C
A
2
+1.5 C
75 cm
Q1
E1 = 11,987 N/C
1
+3.1 C
190 cm
Figuring out direction:
1 = 45o
2 = Tan-1(.75/1.9) = 21.54o
T = 180-21.54 = 158.46o
Really hoping I don’t need to show any more…
Q2