Transcript Topic 9_3__Electric field, potential and energy

Topic 9: Motion in fields
9.3 Electric field, potential and energy
9.3.1 Define electric potential and electric
potential energy.
9.3.2 State and apply the expression for electric
potential due to a point charge.
9.3.3 State and apply the formula relating
electric field strength to electric potential
gradient.
9.3.4 Determine the potential due to one or more
point charges.
9.3.5 Describe and sketch the pattern of
equipotential surfaces due to one and two
point charges.
9.3.6 State the relation between equipotential
surfaces and electric field lines.
9.3.4 Solve problems involving electric potential
energy and electric potential.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Define electric potential and electric potential
energy.
You are probably asking yourself why we are
spending so much time on fields.
The reason is simple:
Gravitational and electric
fields expose the symmetries
in the physical world that are
so intriguing to scientists.
The Physics Data Booklet shows
this symmetry for Topic 6 and
Topic 9:
FYI
Both forces are governed by an
inverse square law.
Mass and charge are the
corresponding physical quantities.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Define electric potential and electric potential
energy.
We define the electric potential energy at a
point P as the amount of work done in moving a
charge q from infinity to the point P.
Without going through a rigorous proof, we recall
that work = force  distance. From Coulomb’s law
F = kq1q2/r2
∆EP = Fr
= (kq1q2/r2)r
= kq1q2/r
electric
∆EP = kq1q2/r or ∆EP = (1/40)q1q2/r
potential
where k = 8.99×109 N m2 C−2
energy
0 = 8.85×10-12 C2 N-1 m−2
FYI
Note the two forms of the proportionality const.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Define electric potential and electric potential
energy.
or EP = (1/40)q1q2/r
electric
EP = kq1q2/r
potential
where k = 8.99×109 N m2 C−2
energy
0 = 8.85×10-12 C2 N-1 m−2
Since at r =  the force is zero, we can
dispense with the ∆EP and consider the potential
energy EP at each point in space as absolute.
EXAMPLE: Find the electric potential energy
between two protons located 3.010-10 meters
apart.
SOLUTION: Use q1 = q2 = 1.610-19 C. Then
EP = kq1q2/r
= (9.0109)(1.610-19)2/ 3.010-10
= 7.710-19 J.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Define electric potential and electric potential
energy.
The potential V (not energy) at any point P in
any field is found in the following way:
Energy needed to bring test
object from infinity to P
potential V =
Test object interacting
with the field
∆V = ∆EP/m
gravitational potential
∆V = ∆EP/q
electric potential
FYI
For the gravitational field the test object is a
mass and the potential is measured in J kg-1.
For the electric field the test object is a
charge and the potential is measured in J C-1.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Define electric potential and electric potential
energy.
From our formula for electric potential energy
∆EP = kQq/r
Note that there is no (-) in
the formula as in V = -Gm/r.
we see that
∆V = ∆EP/q
In the case of the electric
= (kQq/r)/q potential the sign comes
from the charge.
= kQ/r.
V = kq/r
where k = 8.99×109 N m2 C−2
electric
potential
FYI
Again, since at r =  the potential energy is
zero, we can dispense with the ∆V and consider
the potential V at each point in space as
absolute.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State and apply the expression for electric
potential due to a point charge.
V = kq/r
where k = 8.99×109 N m2 C−2
electric
potential
PRACTICE: Find the electric potential at a point
P located 4.510-10 m from a proton.
SOLUTION: q = 1.610-19 C so that
V = kq/r
= (8.99109)(1.610-19)/(4.510-10)
= 3.2 J C-1 (Which is 3.2 V.)
PRACTICE: If we place an electron at P what will
be the electric potential energy stored in the
proton-electron combo?
SOLUTION: From ∆V = ∆EP/q we see that
∆EP = q∆V
= (-1.610-19)(3.2)
= 5.110-19 J (Which is 3.2 eV.)
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State and apply the formula relating electric
field strength to electric potential gradient.
The electric potential gradient is the change in
electric potential per unit distance. Thus the
EPG = ∆V/∆r.
Recall the relationship between the gravitational
potential gradient and the gravitational field
strength g:
g = -∆V/∆r
gravitational potential gradient
Without proof we state that the relationship
between the electric potential gradient and the
electric field strength is the same:
E = -∆V/∆r
electric potential gradient
FYI
In the US we speak of the gradient as the slope.
In IB we use the term gradient exclusively.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State and apply the formula relating electric
field strength to electric potential gradient.
E = -∆V/∆r
electric potential gradient
PRACTICE: The electric potential in the vicinity
of a charge changes from -3.75 V to -3.63 V in
moving from r = 1.80×10-10 m to r = 2.85×10-10 m.
What is the electric field strength in that
region?
SOLUTION:
E = -∆V/∆r
E = -(-3.63 - -3.75)/(2.85×10-10 - 1.80×10-10)
E = -0.120/1.05×10-10 = -1.14×109 V m-1 (or N C-1)
FYI
Maybe it is a bit late for this reminder but be
careful not to confuse the E for electric field
strength for the E for energy!
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Determine the potential due to one or more point
charges.
Since electric potential is a scalar, finding the
electric potential due to more than one point
charge is a simple additive process.
EXAMPLE: Find the electric potential
r
at the center of the circle of protons
shown. The radius of the circle is the
size of a small nucleus, or 3.010-15 m.
SOLUTION: Because potential is a scalar,
it doesn’t matter how the charges are arranged on
the circle. Only the distance matters.
For each proton r = 3.010-15 m. Then each charge
contributes V = kq/r so that
V = 4(9.0109)(1.610-19)/3.010-15
= 1.9106 N C-1 (or 1.9106 V).
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Determine the potential due to one or more point
charges.
EXAMPLE: Find the change in electric
potential energy in moving a proton from
r
infinity to the center of the previous
nucleus.
SOLUTION: Use ∆V = ∆EP/q and V = 0:
∆EP = q∆V
= (1.610-19)(1.9106)
= 3.0 10-13 J
Converting to eV we have
∆EP = (3.0 10-13 J)(1 eV / 1.6 10-19 J)
= 1.9106 eV = 1.9 MeV.
FYI
This is how much work we would need to do to add
the proton to the nucleus.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Describe and sketch the pattern of equipotential
surfaces due to one and two point charges.
Equipotential surfaces are imaginary surfaces at
which the potential is the same.
Since the electric potential
for a point mass is given by
V = kq/r it is clear that the
q
equipotential surfaces are at
fixed radii and hence are
concentric spheres:
equipotential
FYI
surfaces
Generally equipotential surfaces are drawn so
that the ∆Vs for consecutive surfaces are equal.
Because V is inversely proportional to r the
consecutive rings get farther apart as we get
farther from the mass.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Describe and sketch the pattern of equipotential
surfaces due to one and two point charges.
EXAMPLE: Sketch the
equipotential surfaces due
to two point charges.
q
q
SOLUTION: Here is the
sketch of two negative
charges:
Here is the sketch of
two positive charges:
q
FYI
These surfaces are identical to those of two
masses. The difference will be in the electric
dipole (having no gravitational counterpart).
q
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State the relation between equipotential surfaces
and electric field lines.
EXAMPLE: Sketch the equipotential surfaces in the
vicinity of a dipole.
SOLUTION: If you recall that equipotential
surfaces are perpendicular to the electric field
lines the sketches are very simple:
The surfaces are the
dashed ellipses.
FYI
Note that the red Efield is everywhere
perpendicular to the
blue equipotential
surfaces.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State the relation between equipotential surfaces
and electric field lines.
In 3D the point charge
looks like this:
PRACTICE: Explain how the
3D negative point charge
picture would differ from
that of the positive point
charge.
SOLUTION:
Simply reverse the E-field
The positive point
arrows so that they point
charge.
towards the charge instead
of away from it.
The equipotential surfaces would be the same.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State the relation between equipotential surfaces
and electric field lines.
PRACTICE: Sketch the equipotential surfaces and
the E-field surrounding a line of positive
charge.
SOLUTION:
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State the relation between equipotential surfaces
and electric field lines.
PRACTICE: Draw a 3D contour
map of the equipotential
surface surrounding a
negative charge.
SOLUTION:
FYI
If you think of the
equipotential surface
as a hill, a positive
test charge placed on
the hill will “roll”
downhill.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State the relation between equipotential surfaces
and electric field lines.
PRACTICE: Identify this equipotential surface.
SOLUTION:
This is a
dipole.
The positive
charge is the
peak and the
negative
charge is the
pit.
FYI
The E-field
is a
conservative
field just as the g-field is.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
State the relation between equipotential surfaces
and electric field lines.
EXAMPLE: Use the 3D view of the equipotential
surface to interpret the
electric potential
gradient E = -∆V/∆r.
SOLUTION: We can choose
any direction for our r
value, say the
∆y
y-direction:
Then E = -∆V/∆y.
∆V
This is just the
gradient (slope) of
the surface.
Thus E is the (-) gradient
of the equipotential surface.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
Equipotentials are perpendicular to E-field
lines and they never cross each other!
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
From E = -∆V/∆r we see that the closer the Vs
are, the bigger the E.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
+ + + + + + + + + + + + + + +
From conservation of charge, charge can be
neither created nor destroyed.
It can, however, be separated into (+) and (-)
in equal quantities.
Thus, the (-) was separated from the ground,
leaving behind an equal amount of (+).
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
+ + + + + + + + + + + + + + +
The electric field strength E at any point in
space is equal to the force per unit charge at
that point.
If in doubt, find the applicable formula and
translate it into words. “E = F/q.”
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
+ + + + + + + + + + + + + + +
Just recall that between parallel plates the
E-field is uniform (equally spaced), and it
points from (+) toward (-).
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
 = q/A = 20 C / 7106 m2 = 2.8610-6 C m-2.
0 = 8.85×10-12 C2 N-1 m−2 (from Data Booklet).
Then E = /0 = 2.8610-6 / 8.85×10-12 = 3×105 Vm-1.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
 is the same everywhere on the cloud.
0 is the same in air as in free space.
The ground and the cloud’s lower surface
are flat and parallel.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
Use E = -∆V/∆r and ignore the sign:
Then ∆V = E∆r = (3×105)(500) = 1.5108 V!
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
I = q/t = 20 C / 2010-3 s = 1000 A.
∆EP = q∆V = (20)(1.5108) = 3109 J.
Or…P = IV = (1000)(1.5108) = 1.51011 W.
E = Pt = (1.51011)(2010-3) = 3109 J.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
The E-field points from more (+) to less (+).
Use E = -∆V/∆r and ignore the sign:
E = ∆V/∆r = (100 V – 50 V) / 2 cm = 25 V cm-1.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
Electric potential at a point P in space is
the amount of work done in bringing a charge
from infinity to the point P.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
The E-field points toward (-) charges.
The E-field is ZERO inside a conductor.
Perpendicular to E-field, and spreading.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
You can also tell
directly from the
concentration of the
E-field lines.
From E = -∆V/∆r we see that the bigger the
separation between consecutive circles, the
weaker the E-field.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
V = kq/a
V is ZERO inside a
conductor.
V is biggest (-)
when r = a.
From V = kq/r we see
that V is negative and
it drops off as 1/r.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
V = kq/r
V = (9.0109)(-9.010-9)/(4.5 10-2) = -1800 V.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
It will accelerate away from the surface
along a straight radial line.
Its acceleration will drop off as 1/r2 as
it moves away from the sphere.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
q = -1.610-19 C.
V0 = -1800 V.
∆EP = q∆V.
From Vf = kq/r we have
Vf = (9.0109)(-9.010-9)/(0.30) = -270 V.
∆EP = q∆V = (-1.610-19)(-270 - -1800) = -2.410-16 J.
∆EK + ∆EP = 0  ∆EK = -∆EP = 2.410-16 J.
0
EKf – EK0 = 2.410-16 J.
(1/2)mv2 = 2.410-16.
(1/2)(9.1110-31)v2 = 2.410-16.
v = 2.3107 ms-1.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
|E| = ∆V/∆r = (80 – 20) / 0.1 = 600.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.
V = kQ/r
On the xaxis V  0
since r is
DIFFERENT
for the
paired Qs.
At any point on
the y-axis V = 0
since r is same
and paired Qs
are OPPOSITE.
Topic 9: Motion in fields
9.3 Electric field, potential and energy
Solve problems involving electric potential
energy and electric potential.