Potential Difference
By convention, a point in an electric circuit is said to have zero electric potential (potential difference, voltage)
if is grounded (connected to earth). The other way to have zero potential difference is to have a point charge
be located at an infinite distance from another charge.
We have already discussed how to determine the potential difference of a test charge, the change in potential
energy, and the work needed to move the charge. We can now consider the field itself and calculate the
potential difference at a point within the field, regardless of the test charge.
V=kq/d
V is the potential difference at a point that is a distance d from the charge q that created the field.
The potential difference depends only on charge and the distance from the charge.
Superposition Again
When two or more charges are in proximity to one another, the voltage is the sum of the potential difference
from each charge. Potential difference is a scalar; therefore the total electric potential at some point near
several point charges is the algebraic sum of the electric potentials from each of the charges
V = ∑k (q1/d1 + q2/d2 + q3/d3 + …)
q3
q1
r1
r3
r2
q2
Exercise: Find the potential at the point in the center.
V = ∑k (q1/d1 + q2/d2 + q3/d3)
V = ∑ 9•10 Nm / C (0.0015C /0.02m + 0.0025C / 0.027m + 0.002C / 0.018m)
9
2
2
V = 2.5•106V
Remember that voltage is not a vector; it’s a scalar.
Exercise: Calculate the voltage at the top point.
V = ∑k (q1/d1 + q2/d2)
V = ∑ 9•109Nm2 / C2 (5•10-6C / 4m + -2•10-6C / 5m)
V = 7.6•103V
Calculate the work required to bring a 4µC charge to the top point.
The work is equal to the ∆PE.
PE = qV
PE = 4•10-6C (7.6•103 J / C)
PE = 3•10-2J
Velocity and Electric Fields
These calculations can be used to find the speed of an accelerated particle.
Exercise: An electron is accelerated from rest through 120V; calculate its speed.
PE = q V
PE becomes KE
KE = ½ m v2
½ m v2 = q V
v = √(2qV / m)
v = √(2 • 1.6•10-19C • 120V / 9.11-31kg)
v = 5.9•105 m/s
Exercise: What voltage would accelerate an electron need to 1.8 x 107 m/s?
PE = qV
PE becomes KE
KE = ½ m v2
½ m v2 = q V
V = mv2 / 2q
V = 9.11•10-31kg (1.8•107m/s)2 / 2(1.6•10-19C)
V = 922V
Exercise: An electron is 3 cm from the center of a uniformly charged sphere of radius 2 cm. The sphere’s
charge is 1nC. How fast will the electron be traveling when it hits the surface of the sphere?
PE becomes KE
k q1 q2 / r = ½ m v2
v = √(2 k q1 q2 / r m)
9
v = √(2 (9•10 ) (1.6•10-19C • 10-9C) / (0.02m • 9.11•10-31kg)
v = 1.78•107 m/s
Exercise: An electron is fired into at the midpoint of a field between two charged plates at 5.6•106m/s.
The plates are 2mm apart. The ΔV for the plates is 100V.
(a) Find the magnitude of the electric field.
(b) Make a drawing of the two plates using arrows to show the direction of the field
(c) Calculate the force the electron experiences.
(d) Calculate the electron’s acceleration.
(e) Determine when the electron will hit the upper plate.
(f) Determine where the electron will hit the upper plate.
(a)
E=V/d
E = 100V / 0.002m
E = 50,000 V/m
(b)
The electric field shows what happens to a positive charge.
If a negative charge goes up, a positive charge would go down.
(c)
F = qE
F = 1.6•10-19C • 50,000 N/C
F = 8•10-15N
(d)
a = F /m
a = 8•10-15N / 9.11•10-31kg
a = 8.78•1015 m/s2
(e)
t = √ (2d / a)
t = √ (2•10-3m / 8.78•1015 m/s2)
t = 4.77•10-10s
(f)
x = vt
x = 5.6•10-6m/s • 4.77•10-10s
x = 2.7mm
Equipotential Curves:
The lines of force for a positively charged conducting sphere look like this.
An “equipotential curve” is a continuous path that follows a single potential, it traces a shape that has the
same potential difference at every point. The equipotential curve for the charged sphere would be another
sphere (a circle on this two dimensional drawing) because the voltage depends only on the distance from the
charge. Curves nearest the surface have greatest potential.
Potential difference is given by V = kq / r
As the distance (r) increases, the potential difference decreases. So equipotential curves that represent a
specific voltage difference (i.e. each curve represents a decrease of 5 volts) would look like this.