
Electrical Potential Energy:
◦ Potential energy associated with a charge due to its
position in an electric field.

ME = KE + PEgrav + PEelastic + PEelectric

Electrical potential energy in a uniform
electric field:
◦ PEelectric = -qEd
◦ q = charge (Coulomb - C)
◦ E = electric field strength (Newton/Coulomb – N/C)
◦ d = displacement in the direction of the electric field (m)
Will the particle shown gain or
lose PEelectric as it moves to the
right?
Will it gain or lose PEelectric as it
moves up? Down?
If the particle started at point
B, will it gain or lose PEelectric as
it moves to the left?

Potential Difference in a Uniform Electric Field
◦ The work that must be exerted against electric
forces to move a charge a certain displacement,
divided by the charge.
PEelectric
V 
q
qEd
V 
q
V  Ed

Sample Problem
◦ A charge moves a distance of 2.0 cm in the direction of a
uniform electric field whose magnitude is 215 N/C. As the
charge moves, its electrical potential energy decreases by 6.9 
10-19 J. Find the charge on the moving particle. What is the
potential difference between the two locations?

Answers:
◦ q =1.6 * 10-19 C
◦ ΔV = -4.3 V
The electric potential at point
A depends on the charge at
point B and the distance
between them, r.
Similarly, the electric
potential at point B depends
on the charge at point A and
the distance between them, r.

Usually we compare the electric potential of a
point to that of a point infinitely far away.
This is what the potential difference is
referring to.
q
q
V  kC  kC
r

q
V  kC  0
r
q
V  kC
r
Find the electric potential at the black point in the
middle of the diagram. (kC = 8.99 * 109 Nm2/C2)
Answer : ΔV = -5.1 * 104 V

Homework:
◦ Pg. 585 # 1-3

Capacitor
◦ Device used to store electrical potential energy.
◦ Made up of two conducting plates oriented parallel
to each other.

Capacitance
◦ Ability of a conductor to store energy in the form of
electrically separated charges.

Capacitance is the ratio of charge to
potential:
Q
C
V

The unit for capacitance is Farad (F)
◦ A Farad is equal to Coulomb per Volt (C/V)


Sample Problem:
◦ A 225 F is capacitor connected to a 6.00 V battery
and charged. How much charge is stored on the
capacitor?

Answer:
◦ Q = 1.35 * 10-3 C

Capacitance depends on three characteristics
of the metal plates:
◦ Size of the plate
◦ Distance between the plates
◦ Material between the plates
A
C  0
d




C – Capacitance (F)
A – Area of plate (m2)
d – Distance between plates (m)
ε0 – Permittivity of a vacuum
◦ 8.84 * 10-12 C2/N*m2
A
C  0
d

Dielectric
◦ The material in between the plates of a capacitor
◦ Adding a dielectric increases the capacitance and
reduces the electric field between the plates
◦ Changing the material changes the variable ε0

Sample Problem:
◦ A parallel plate capacitor has an area of 2.0 cm2,
and the plates are separated by 2.0 mm. What is
the capacitance (assume it is a vacuum between the
plates)?

Answer:
◦ 4.8 * 10-6 F

The potential energy that a capacitor stores
depends on:
◦ Charge on only one of the plates
◦ Potential difference between the plates
1
PEelectric  QV 
2


Sample Problem
◦ A capacitor, connected to a 12 V battery, holds 36
µC of charge on each plate. What is the capacitance
of the capacitor? How much electrical potential
energy is stored in the capacitor?

Answers:
◦ C = 3 * 10-6 F
OR
◦ PEelectric = 2.16 * 10-4 J
C = 3 μF

Electric Current (I)
◦ The rate at which electric charge passes through a
given area.
Q
I
t

The unit of I is Coulombs per second, or
Amperes (A).
◦ Sometimes abbreviated Amps


Resistance to Current
◦ Opposition to the flow of charge
 Similar to friction when dealing with motion

Ohm’s Law
◦ The unit for resistance volt per ampere, or ohm (Ω)
V
R
I
V  IR

Ohm’s Law
◦ Applies to materials where the resistance is
constant over a wide range of potential differences.
 Ohmic material
◦ When the resistance of material is not constant over
a wide range of potential differences, this material
does not obey Ohm’s Law.
 Non-Ohmic material

Sample Problems
◦ A typical 100 W light bulb has a current of 0.83 A.
How much charge flows through the bulb filament
in 1.0 h? How many electrons would flow through in
the same time period?
 Answers: 3.0 * 103 C, 1.9 *1022 electrons
◦ This same 100 watt bulb is connected across a 120
V potential difference. Find the resistance of the
bulb.
 Answer: 140 Ω


Varying the resistance can vary the current
through a circuit with a certain potential.
Applications
◦ The human body has a resistance value between
500,000 Ω (dry) to 100 Ω (soaked in salt water).
◦ 0.01 A can cause tingling in the body
◦ 0.15 A can cause disruptions in the heart

Homework

Batteries and generators supply energy.

Current can flow in two different forms:
◦ Direct current
 Charge moves in a single direction
◦ Alternating Current
 Direction of current continually changes

Energy Transfer
Is the electrical potential
energy gained, lost, or unchanged
as the electrons flow through the
following portions of the circuit shown:
A to B?
B to C?
C to D?
D to A?

And when we graph the potential energy, we
get a graph that looks like this:

Electric Power
◦ Rate of energy consumption.
P  IV

The units for power are Amps * Volts, or
Watts (W)
◦ Watts are also a unit for Energy per unit time (J/s)



We can manipulate this Power equation using
Ohm’s Law.
Power dissipated by a resistor:
P  IV & V  IR
V
P  I R&P 
R
2
2

Sample Problems
◦ A toaster is connected across a 120 V kitchen
outlet. The power rating of the toaster is 925 W.
 What current flows through the toaster?
 What is the resistance of the toaster?
 How much energy is consumed in 75.0 s?

Answers
◦ 7.7 A
◦ 16 Ω
◦ 6.94 * 104 J

Homework