Stream lines . front plate slightly charged induces opposite charge on back plate. Brushes pull off charges charges collected in leyden jar (capacitor). Wimshurst Machine. Electric Field. Definition . Electric field is the
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Wimshurst Machine
strength & direction of the forces in space surrounding a positive test charge
rewrite the force equation
F = q1 * E
q1 is the positive test charge
E is the field created by charge(s) q2
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C
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Electric Fields
Electric Field between 2 unlike charges
Note that the distance between
electric field lines at C is shorter
than that at B.
Distance between electric field
lines at B is shorter than at A.
Electrostatics
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Electric Fields
Electric Field between 2 charged plates
Electrostatics
Charles Coulomb~1750
electron potential energy (PEe or U)
High energy: + is close to +: ++
or + is far from –: + 
High energy work needed to move + charge
(W = F*d)
E
F
+
low energy
(a) When a positive charge moves in the direction of an electric field, the field does positive work and the potential energy decreases. Work = qo E d
POSITIVE charge moving in an E field.
1 joule of energy 9x104 joule energy
0.001 C of charge 1x104C of charge
1/.0001 = 1000 volts 9/1 = 9 Volts
1,000,00 volts (high) 1 joule of energy (low) .00001 coulombs of charge
1,000,000 v = 1 j/ .00001c
Uniform electric field electric field, the 2 situations Point Charges
Force: F=q*E Force= k q1*q2 / d2
Electric potential energy Electric potential energy
PE= F*d = q*E*d PE = F*d = k q1*q2 / d
Voltage (electric potential)Voltage (electric potential)
V= PE/q1 = q*E*d /q = E*d V= PE/q1 = k q2 / d2
Voltage difference: Voltage difference:
DVe = D PE/q1 = E* Dd DVe = D PE/q1 = k q2 / d2  kq2/ d1
work
Electrostatics takes work to lift against gravitygravitational
Force: F= kq1 q2 / d2 F= Gm1m2/d2
Field E= F/q1 = kq2/d2 g = F/m1 = Gm1/d2
potential energy PE= F*d = kq1 q2 / d PE = F*d= Gm1m2/d
(using field) =qE*d = mg*h
Potential V= PE/q1 = kq2 / d U= PE/m1 = G m2 / d
(Using field) = E*d = g*h