Physics Class Notes
Electric Potential Energy
Electric Potential (voltage) (V)
Electric Potential Difference (∆V)
Potential Energy W = Fd
= q+ Ed
Eelectric field = F/q+
F =q+ E
q+ = positive test charge
Electrical energy of a charge is equal to the amount of WORK needed to move a charge across an
electric field. This energy can be stored on the charge as POTENTIAL ENERGY or retrieved from the charge
as WORK.
WORK = PE or KE
 PE = -W
 PE = qEd
this is only a true equation when
E = uniform electric field
W = qEd
q = + or - charge
E = strength of electric field
+
d = distance moved through electric field
= + or - depending if the charge moves
with, or against the field
High Potential
(high voltage)
relative to electric field
Low Potential
(low voltage)
A Uniform Electric Field is set up between two oppositely charged plates that are close together
Example A
How much potential energy is found in a +12mC charge in an electric field of 1500N/C when it is:
PE = qEd
a. 25 cm from the negative plate (4.5J)
b. 15 cm from the negative plate (2.7J)
How much work is required to move the +12mC charge from the 15 cm position to the 25 cm position?
Work of electric field= ∆PE = (PEposition 2 – PEposition 1)
= (4.5J – 2.7J) = 1.8J (positive work because a + charge moved against the
electric field) energy was added to move charge
How much PE is acquired by the +12mC charge as it moves from the 15 cm position to the 25 cm position?
Work = +∆PE (1.8J) = + 1.8 J of PE
How much work is required to move the +12mC charge from the 25 cm position to the 15 cm position
Work = ∆PE = (PEposition 2 – PEposition 1)
= (2.7 – 4.5) = -1.8J (positive work gained because a + charge moved with the
electric field)
How much PE is acquired by the +12mC charge as it moves from the 15 cm position to the 25 cm position?
Work = +∆PE (-1.8J) = - 1.8 J of PE
ELECTRIC POTENTIAL or VOLTS (V) The potential energy on a charge (q) due to its position in a
uniform electric field.
V = PE/q = qEd/q = Ed
Volt =Ed
V = Ed
potential energy available to a charge
electric potential = Ed in a uniform electric field
electric potential = Ed between parallel plates only
Example B
Find the electric potential (V) for the 12mC charge in Example A when it is at the 25cm position.The
electric field between the plates is 1500N/C
V = PE/q
V = 4.5J/ 12x10-3C) = 375 J/C = 375 volts = 375V
or
V = Ed = 1500N/C( 0.25m) = 375Nm/C = 375 J/C = 375 V
Equation for electric potential when electric field is uniform
Example C
What is the electric potential for a charge placed in a 1500N/C electric field 45cm from the negative
plate?
electric potential = Ed = 1500N/C(0.45m) = 675Nm/C = 675V
When determining the position of a charge in an electric field the negative plate is considered the zero position,
the low potential position. The positive plate is the high potential position
ELECTRIC POTENTIAL DIFFERENCE (ΔV) (VOLTAGE)
Electric potential difference is the change in available potential energy as a charge moves across an electric
field
ΔV = (Vb-Va) = (Edb – Eda) = ΔPE/q = Δ(qEd)/q = ΔEd
ELECTRIC POTENTIAL DIFFERNCE between parallel plates (uniform electric field)
ΔV = (Vb-Va) = (Edb – Eda) = ΔPE/q = Δ(qEd)/q = ΔEd
As a charged particle moves to a new position relative to the plates generating the electric field, the amount of
PE available to the charge will vary according to its position.
Example D
Find the potential difference between the stated points in the following uniform electric field.
Between the 0.35m position and the 1.35m position in a uniform electric field of 1800N/C
ΔV = (Edb – Eda) = (1800N/C∙0.35m -1800N/C∙1.35m)
ΔV = 630V – 2430V = -1800 V = -1800J/C
The positively charged plate is the high potential, the negatively charged plate is the low potential.
(a) Moving a proton from the negative to the positive plate increases the proton’s potential energy.
(b) When it is released from the positive plate, the proton accelerates toward the negative plate,
gaining kinetic energy at the expense of electric potential energy.
(c) It takes the same work to move A to B as A to B’
The work done to move a proton between any two points in an electric field, such as A and B or A
and B’ is independent of the path.
Example E
An electron is accelerated through a potential difference of +4,500V (goes towards high potential) . Determine
the final velocity of the electron if it was initially at rest. ( mass of electron = 9.1 x 10-31kg)
vf = ?
vo = 0.0 m/s
ΔV = + 4,500V
ΔKE = -ΔPE
ΔPE = -7.2 x 10-16J
Δ V = ΔPE/C
ΔPE = qEΔd = q(ΔV)
ΔPE = -ΔKE
ΔKE = ½ mv2
7.2 x 10-16J = ½ (9.1 x 10-31kg)v2
ΔPE = (-1.6 x 10-19C) (+4,500J/C)
-7.2 x 10-16J = - ΔKE
ΔKE = + 7.2 x 10-16J
v2 = 1.58 x 1016 m2/s2
v = 3.98 x 107m/s
ELECTRIC POTENTIAL DIFFERNCE due to a point charge
The electric field around a point charge is not uniform. The electric field weakens as gets further from the
positive charge.
When analyzing values of potential difference for a point charge consider the following:
 Electric potential increases when moving closer to a positive point charge and when moving
further from a negative charge
 Electric potential decreases when moving further from a positive charge and when moving
closer to a negative charge.
V = Ed = (kQ/r2)r = kQ/r
ΔV = kq/rb – kq/ra
(for point charge only)
+q
ra
rb
ΔV = Vb –Va = kq/rb – kq/ra
Example F
What is the electric potential difference moving between two points 5.0m and 7.0m from a point charge
of +6.0mC?
ΔV = Vb –Va = kq/rb – kq/ra
ΔV = kq/rb – kq/ra = 9x109Nm2/C2(6.0 x 10-3C)/7m - 9x109Nm2/C2(6.0 x 10-3C)/5m
= 7.71 x 106V- 1.08 x 107V = -3.09 x 106V
= -3.1 x 106V
Example G
What is the electric potential difference moving between two points 8.0m and 4.0m from a point charge
of +3.0nC?
ΔV = Vb –Va = kq/rb – kq/ra
ΔV = kq/rb – kq/ra = 9x109Nm2/C2(3.0 x 10-9C)/4m - 9x109Nm2/C2(3.0 x 10-9C)/8m
= 6.75V – 3.38V = +3.37V
= +3.4V