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Electrostatics
Overview
Electrostatics - Concepts
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Definition
•
•
Electrostatics = Electron + stationary
“The study of electric charges, forces and fields”
Symbol for charge
•
Letter “q” or “Q”
Units of charge
•
Coulombs (C) - standard
• 1 C = the charge on 6.25 x 1018 electrons
• 1 electron has a charge of 1.6 x 10-19 C
• microCoulombs (µC) - common
• X 10-6 (ex. 3 x 10-6 C = 3 µC)
Charges
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Charges can be either
• Negative (-)
•
an excess of electrons
• Positive (+)
• a deficiency of electrons
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Fundamental or elementary charge
• Electron
• “Quantized” – integer numbers
Atomic Structure
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Electrons are free to move,
but not protons or neutrons
Positive and negative
particles, when brought
close together are subject
to the following “BIG
RULE”:
• Like (+/+ or -/-) charges
•
repel
Unlike (+/-) charges attract
Overall – charge is conserved
Electrostatic Force - Coulomb’s
Law
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Two charged particles q1 & q2
interact with each other with a
force that is:
•
Proportional to the product of
their charges, and
• Inversely proportional to the
square of their separation
distance (d)
• where
• k = electrostatic constant
• k = 9 x 109 N.m2/C2
 (Cf Newton’s Law of Universal
Gravitation)
Practice – Coulombs’ Law
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What is the magnitude of the e-static force
between an electron and proton of
hydrogen, when they are separated 5.3 x
10-11 m? (qe = -1.6 x 10-19 C, qp = 1.6 x 10-19
C)
Solve Coulomb’s Law for F
• F = k(q1q2)/d2
• F = 9 x 109 x (-1.6 x 10-19 C) x (1.6 x 10-19 C)
(5.3 x 10-11)2
• F = 8.2 x 10-8 N
Practice – Coulomb’s Law
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
Two e-static charges of 60 x 10-6C and
50 x 10-6C exert a force of 175N on each
other. What is their separation distance?
Solve Coulomb’s Law for d:
• F = k(q1q2)/d2
• 175 = 9 x 109 x (60 x 10-6)(50 x 10-6)
• d2 = 1.54 x 10-1
• d = 3.93 x 10-1 m
d2
Electric Field (E)
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An electric field is a region
in space in which an
electric charge experiences
a force.
Surrounding electric
charges are lines of force
which form an electric
field. These lines of force
flow out of a positive
charge and into a negative
charge, as shown.
Represented by symbol E
E is a vector!
E
Electric Field Strength?
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The Electric Field strength can be
determined by 2 methods:
Method 1: The strength of this electric
field (E) at a point is equal to the force
felt by a small positive test charge (q0)
being placed at that point.
• E = F/q0
 Newtons/Coulomb (N/C)
Electric Field Strength?
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Method 2: the electric field strength at a point P, d units
distant from the charge of interest q is equal to the
product of the charge and the electrostatic constant (k)
divided by the square of the separation distance, or
• E = kq/d2
(N/C) 
Practice – Electric Field

A force of 2.1 N is exerted on a 9.2e-4 C
test charge when it is placed in an electric
field created by a 7.5 C charge. If the
force is pushing it West,
• determine the electric field (E) at that point.

Solve E = F/q0 for E
• E = 2.1/9.2e-4
• E = 2282.6 N/C
Practice – Electric Field
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If a positive test charge of 3.7e-6 C is put in
the same place in the electric field as the
original test charge in the last example,
• determine what force (F) will be exerted on it.

Solve E = F/qo for F
• 2282.6 = F/3.7e-6
• F = 8.44e-3 N
Practice – Electric Field
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A tiny metal ball has a charge of –3.0e-6 C.
• determine the magnitude and direction of the
field at a point, P, 30cm away?

Solve E = kq/d2 for E
• E = (9e9)(-3e-6)/0.32
• E = -300,000 N/C
Electrical Potential Difference
(Voltage V)
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Definition
• The amount of work done
(change in electrical
potential energy) in moving
a +ve test charge (q0) from
one location to another in an
electrical field (E).
• V = W/q0
 in
joules/coulomb = volts
• If the charge is moving
against the field, then the
work done is negative, and
vice versa.
Practice – Potential Difference
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Determine the electric potential
difference (or voltage V) of a 3.4 C
charged object (q) that gains 2.6 x 103 J
as it moves through an electric field.
Solve V =W/q for V
• V = (2.6x 103)/3.4C
• V = 7.6 x 102 V
Parallel Plates
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If an electric field (E) exists
between 2 parallel plates, then the
field is uniform except at the
edges.
The electric field depends on the
potential difference (V) across the
plates and the plate separation (d),
as follows:
• E = V/d,
• units of volt/meter (equivalent to
N/C)
Practice – Parallel Plates
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If we have two parallel plates that are
16.0mm apart and want a uniform field of
800 N/C between these plates, determine
the voltage we must apply to the plates.
Solve E = V/d for V
• V = (800N/C) (0.0160m)
• V = 12.8 V