Principles of Technology
Chapter 9 Static Electricity 3
Name_______
KEY OBJECTIVES
At the conclusion of this chapter you will be able to:
• Define the terms potential difference and electric potential, and state the SI units for measuring these
quantities.
• Define the term electron-volt and relate it to the joule.
• Relate the electric field strength between oppositely charged parallel plates to the potential difference
across them, and use this relationship to solve problems.
• Describe Millikan’s oil drop experiment and its contribution to the u of electric charge.
• Define the term capacitance and state its SI unit.
• Solve simple capacitance problems.
9.10 POTENTIAL DIFFERENCE
We have described the electric field in terms of the force on a charged particle. We can also describe the
electric field in terms of work and energy as shown in the diagram.
In the diagram, we move a test charge q
between two points, A and B, in an electric
field. If the charge is repelled by the field,
we must do work on the charge to move it
between the two points. The work we do
against the field (WAB) will increase the
potential energy of the test charge.
Another way of describing this situation is
to say that a potential difference exists
between points A and B in the electric field.
We define this potential difference (V) as follows:
V=W/q
Potential difference is a scalar quantity, as is work. The unit of potential difference is the joule per
coulomb (J/C) called the volt (V) in honor of Alessandro Volta, an Italian scientist.
PROBLEM
When a charge of -4 x 10-3 coulomb (q) is moved between two points in an electric field, 0.8 joule (W)
of work is done on the charge. Calculate the potential difference (V) between the two points.
SOLUTION
The sign of the charge is not needed to solve the problem. We need know only the magnitude of the
charge and the work done on it.
V=W/q
V = 0.8 J / -4 x 10-3 C
V = 200 V
Assessment Question 1
When a charge of 85 coulombs (q) is moved between two points in an electric field, 5200 joules (W) of
work is done on the charge. Calculate the potential difference (V) between the two points. (V=W/q)
a) 0.16 V
b) 61 V
c) 435 V
d) 5300 V
PROBLEM
Calculate the work (W) done on an elementary charge (q) that is moved between two points in an
electric field with a potential difference of 1 volt (V).
SOLUTION
W=qV = (1.6 x 10-19 C) (1.0 V) = 1.6 x 10-19 J
Assessment Question 2
If a charge of 0.21 coulombs (q) is moved between two points in an electric field, with a potential
difference of 630 V (V), calculate the joules (W) of work is done on the charge
between the two points. (W=qV)
a) 0.33 V
b) 66 V
c) 132 V
d) 300 V
The very small quantity of work in the problem shown above is frequently used
as a unit of energy in atomic and nuclear physics. This unit is known as an
electron-volt (eV).( 1 eV = 1.6 x 10-19 J) Some multiples of the electron-volt are
shown in the table, where the letter M stands for mega-; the letter G for giga-; and the letter T for tera.
PROBLEM
A charge, equal to 2 x 107 elementary charges (q), is moved through a potential difference of 3000
volts (V). What is the change in the potential energy of the charge?
SOLUTION
The change in the potential energy of the charge is equal to the work done on it. We can calculate this
work directly in electron-volts by using the unit elementary charge, rather than coulomb, for the charge:
W = q V = (2 x l07)(1.6 x 10-19 C)(3000 V) =6x l010 eV = 60GeV
Assessment Question 3
A charge, equal to 5 x 104 elementary charges (q), is moved through a potential difference of 120 volts
(V).
1 elementary charge = 1.6 x 10-19 C
What is the change in the potential energy (W) of the charge.
(W=qV)
a) 6 V
b) 6 eV
c) 6 MeV
d) 6 GeV
9.11 ELECTRIC POTENTIAL
We know that it is possible to measure the potential
difference between two points, A and B, by dividing
the work done on a charge by the magnitude of the
charge. Suppose, however, that we wish to know the
electric potential at just one of the points, A or B. How
can we accomplish this?
This situation is similar to climbing 525 meters
vertically between two points on a hill. It seems
only natural to ask how high the first and second
points are.
The secret is to establish a reference point whose
value is zero. With regard to the hill, sea level, with
a height of 0 meter, is the reference point, and the
height (or altitude) of each point is the distance of
that point above sea level. The distance between
two points is then the difference in their altitudes. The diagram above illustrates this concept.
Similarly, to assign electric potentials, we establish a
reference point of 0 volt. For an isolated charge, the
reference point is taken to be infinitely far from the charge.
For other situations, the ground may be taken as a
reference point. We then measure the potential difference
between the point in question and the reference point,
assigning this value as the electric potential of the point. The diagram above, where A represents the
point in question, illustrates how this is accomplished. The electric potential at a point is defined as the
work needed to move a charge of +1 coulomb from infinity to the point in question.
Assessment Question 4
Which set of statements is true:
i. If a charge is repelled by a field, we must do work on the charge to move it between the
two points.
ii. 1 MeV> 1 GeV > 1 TeV
iii. To assign electric potentials, we establish a reference point of 0 volt.
iv. For an isolated charge, the reference point is taken to be adjacent to the charge.
v. For situations other than an isolated charge, the ground may be taken as a reference point.
a) i , iv, v
b) i , ii, iv
c) ii , iii, v
d) i , iii, v
9.12 FIELD STRENGTH AND POTENTIAL DIFFERENCE
Both field strength and potential difference measure qualities of an
electric field, and so it seems only natural that they should be related.
We will establish this relationship for a very simple field
configuration: oppositely charged parallel plates.
The diagram shows a pair of oppositely charged parallel plates a
distance d apart. The potential difference between the plates is V, and
a charge of +q is to be moved from the negative to the positive plate
against the electric field (E) at constant velocity. To do this, we must
supply a force (Fapplied) that is exactly equal to the force that the electric field places on the charge.
The work (W) that we do in moving the charge is (Fapplied), which is also equal to V∙ +q. The force placed
on the charge by the field (F) is E∙ +q. We can combine and simplify all of these relationships as shown
below:
W = (Fapplied)d = V∙+q
F = E∙+q
Combining the three relationships gives:
V / d = F / +q = E
We have now produced the relationship between field strength and potential difference: E = V / d
When this equation is used, the unit for electric field is the volt per meter (V/m), which is equivalent to
the unit newton per coulomb.
Assessment Question 5
If a charge of 0.33 coulombs (q) is moved between two points in an electric field separated by a
distance of 0.6 m (d), with a potential difference of 990 V (V), calculate the force (F) on the charge
between the two points. (F=qV/d)
a) 30 N
b) 180 N
c) 196 N
d) 545 N
PROBLEM
Calculate the uniform electric field (E) between two parallel plates if the potential difference between
them is 50. volts (V) and they are 2.5 millimeters (d) apart.
SOLUTION
To solve the problem, we need to use the relationship E= V/d. First, however, we must convert the
given distance to meters: 2.5 mm ∙ (1 m / 1000 mm) = 2.5 x 10-3 m
Now we can substitute in the equation and do the computation:
E = V / d = 50 V / (2.5 x 10-3 m) = 2.0 x 104 m
Assessment Question 6
Calculate the uniform electric field (E) between two parallel plates if the potential difference between
them is 350. volts (V) and they are 0.7 meters (d) apart.
(E=V/d)
a) 0.2 V/m
b) 17 V/m
c) 357 V/m
d) 500 V/m
Assessment Question 7
Calculate the distance (d) between two parallel plates if the potential difference between them is 770
volts (V) and they are in a uniform electric field with strength of 4.9 V/m (E).
(E=V/d)
a) 0.06 m
b) 17 m
c) 157 m
d) 400 m
9.13 THE MILLIKAN OIL DROP EXPERIMENT
Robert Millikan, an American physicist, used a modified
pair of charged parallel plates to measure the charges on
microscopic oil droplets. The diagram shows how the
observation was performed.
The charged droplet was injected through the opening
into the uniform field between the parallel plates, where
it was observed by a microscope. By changing the
potential difference between the plates, the electric field
strength was varied until the upward electric force on the
droplet was balanced by the weight of the droplet. Millikan was then able to calculate the electric
charge on each oil droplet that he observed. By measuring thousands of such charges, he determined
that the charges were all multiples of 1.60 x 10-19 coulomb and thus concluded that the smallest
charge, the elementary charge, is equal to 1.60 x 10-19 Coulomb.
9.14 CAPACITANCE
When a pair of uncharged parallel conducting plates is connected to a source of potential difference, the
plates begin to charge and they continue charging until the potential difference between them is equal to
the potential difference of the source. These oppositely charged plates constitute a device known as a
parallel plate capacitor. Capacitors store charge and energy, which may be used in applications
such as the flash attachment of a camera. When the flash is turned on, the capacitor builds up charge.
As a picture is taken, the flash is triggered by the discharging capacitor.
The charge that the capacitor accumulates depends directly on the potential difference, and we
can write this equation:
q = CV
The constant of proportionality, C, is known as the capacitance. The more capacitance a capacitor
has, the more charge it will store for a given potential difference. The capacitance of a parallel plate
capacitor depends on the area of the plates, the distance between the plates, and the material present
between the plates. The SI unit of capacitance is the coulomb per volt (C/V) which is known as a
farad (F) in honor of English chemist and physicist Michael Faraday. The farad is a very large unit,
and for most practical purposes the range of capacitance extends from 10-12 farad (a picofarad, pF) to
10-6 farad (a microfarad, μF).
PROBLEM
A pair of parallel plates with a capacitance of 3.0 x 10-6 farad (C) is charged using a 12-volt (V) battery.
Calculate the charge (q) on each plate.
SOLUTION
q = CV =(3.0 x 10-6 F)(l2V) = 3.6 x 10-5 C
Assessment Question 8
A pair of parallel plates with a capacitance of 7.0 x 10-2 farad (F) is charged using a 9-volt (V) battery.
Calculate the charge (q) on each plate.
(q = CV)
a) 0.079 C
b) 0.63 C
c) 16 C
d) 67 C
Assessment Question 9
Calculate the capacitance (C) between two parallel plates if the potential difference between them is
0.950 volts (V) and the charge on each plate is 850 coulombs (q). (q = CV)
a) 0.1 F
b) 1.7 F
c) 807 F
d) 895 F
Assessment Question 10
Which set of statements is true:
i. All electrical charges are multiples of 1.60 x 10-19 coulomb
ii. There are charges smaller than 1.60 x 10-19 Coulomb, known as elementary charges.
iii. When the flash of a camera is turned on, the capacitor builds up charge and the flash is
triggered by the discharging capacitor..
iv. The capacitance of a parallel plate capacitor depends on the area of the plates, the
distance between the plates, and the material present between the plates.
v. 1farad < 1 microfarad < 1 picofarad
a) i , iv, v
b) i , iii, iv
c) ii , iii, v
d) i , iii, v