week 4
Which requires the most
work, to move a positive
charge from P to points 1, 2,
3 or 4 ? All points are the
same distance from P.
1) P → 1
2) P → 2
3) P → 3
4) P → 4
5) all require the same
amount of work
3
2
1
P
4
Which requires the most
work, to move a positive
charge from P to points 1, 2,
3 or 4 ? All points are the
same distance from P.
For path #1, you have to push the
positive charge against the E field,
which is hard to do. By contrast,
path #4 is the easiest, since the
field does all the work.
1) P → 1
2) P → 2
3) P → 3
4) P → 4
5) all require the same
amount of work
3
2
1
P
4
1) P → 1
Which requires zero work, to
move a positive charge from
P to points 1, 2, 3 or 4? All
points are the same distance
from P.
2) P → 2
3) P → 3
4) P → 4
5) all require the same
amount of work
3
2
1
P
4
Which requires zero work, to
move a positive charge from
P to points 1, 2, 3 or 4? All
points are the same distance
from P.
1) P → 1
2) P → 2
3) P → 3
4) P → 4
5) all require the same
amount of work
For path #3, you are moving in a
direction perpendicular to the field
lines. This means you are moving
along an equipotential, which
requires no work (by definition).
Follow-up: Which path requires the least work?
3
2
1
P
4
The positive charge is the end view
of a positively charged glass rod. A
negatively charged particle moves in
a circular arc around the glass rod.
Is the work done on the charged
particle by the rod’s electric field
positive, negative or zero?
1. Positive
2. Negative
3. Zero
The positive charge is the end view
of a positively charged glass rod. A
negatively charged particle moves in
a circular arc around the glass rod.
Is the work done on the charged
particle by the rod’s electric field
positive, negative or zero?
1. Positive
2. Negative
3. Zero
1. the field does positive work on it and
the potential energy increases
When a positive charge
moves in the direction of
the electric field,
2. the field does positive work on it and
the potential energy decreases
3. the field does negative work on it and
the potential energy increases
4. the field does negative work on it and
the potential energy decreases
1. the field does positive work on it and
the potential energy increases
When a positive charge
moves in the direction of
the electric field,
2. the field does positive work on it and
the potential energy decreases
3. the field does negative work on it and
the potential energy increases
4. the field does negative work on it and
the potential energy decreases
The work done by the field is
dW = qE ⋅dr
In this case the vectors E and dr are parallel
and the charge is positive so
dW >0
The change in potential energy is the negtive of
the work done by the field, so
ΔU = - W <0
1. the field does positive work on it and
the potential energy increases
When a negative charge
moves in the direction
of the electric field,
2. the field does positive work on it and
the potential energy decreases
3. the field does negative work on it and
the potential energy increases
4. the field does negative work on it and
the potential energy decreases
1. the field does positive work on it and
the potential energy increases
When a negative charge
moves in the direction
of the electric field,
2. the field does positive work on it and
the potential energy decreases
3. the field does negative work on it and
the potential energy increases
4. the field does negative work on it and
the potential energy decreases
The work done by the field is
dW = qE ⋅dr
In this case the vectors E and dr are parallel
and the charge is negative so
dW <0
The change in potential energy is the negtive of
the work done by the field, so
ΔU = - W >0
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. Which
feels the larger electric
force?
1) proton
2) electron
3) both feel the same force
4) neither – there is no force
5) they feel the same magnitude
force but opposite direction
electron
electron
+
proton
proton
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. Which
feels the larger electric
force?
1) proton
2) electron
3) both feel the same force
4) neither – there is no force
5) they feel the same magnitude
force but opposite direction
Since F = qE and the proton and electron
have the same charge in magnitude, they
both experience the same force. However,
the forces point in opposite directions
because the proton and electron are
oppositely charged.
electron
electron
+
proton
proton
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. Which
has the larger acceleration?
1) proton
2) electron
3) both feel the same acceleration
4) neither – there is no acceleration
5) they feel the same magnitude
acceleration but opposite direction
electron
electron
+
proton
proton
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. Which
has the larger acceleration?
1) proton
2) electron
3) both feel the same acceleration
4) neither – there is no acceleration
5) they feel the same magnitude
acceleration but opposite direction
Since F = ma and the electron is much less
massive than the proton, then the electron
experiences the larger acceleration.
electron
electron
+
proton
proton
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. When it
strikes the opposite plate,
which one has more KE?
1) proton
2) electron
3) both acquire the same KE
4) neither – there is no change of
KE
5) they both acquire the same KE
but with opposite signs
electron
electron
+
proton
proton
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. When it
strikes the opposite plate,
which one has more KE?
1) proton
2) electron
3) both acquire the same KE
4) neither – there is no change of
KE
5) they both acquire the same KE
but with opposite signs
Since PE = qV and the proton and electron
have the same charge in magnitude, they
both have the same electric potential energy
initially. Because energy is conserved, they
both must have the same kinetic energy after
they reach the opposite plate.
electron
electron
+
proton
proton
Here is a simple problem we already know how to solve using forces and F=ma.
But let’s solve it using conser vation of energy:
The electric field strength is 50,000 N/C inside a parallel plate capacitor with a
2.0 mm spacing. A proton is released form rest at the positive plate. What is the
proton’s speed when it reaches the negative plate?
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Ans: 1.38 x 10 m/s
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• 1. Ua = Ub = Uc = Ud
Rank in order, from largest to
smallest, the potential energies
Ua to Ud of these four pairs of
charges. Each + symbol
represents the same amount of
charge.
• 2. Ud = Ub > Uc = Ua
• 3. Ud > Ub = Uc > Ua
• 4. Ub > Ua > Ud > Uc
• 5. Ud > Ub = Uc > Ua
• 1. Ua = Ub = Uc = Ud
Rank in order, from largest to
smallest, the potential energies
Ua to Ud of these four pairs of
charges. Each + symbol
represents the same amount of
charge.
• 2. Ud = Ub > Uc = Ua
• 3. Ud > Ub = Uc > Ua
• 4. Ub > Ua > Ud > Uc
• 5. Ud > Ub = Uc > Ua
q2
Ua = k
r
2q 2
Ub = k
r
(2q)q
q2
Uc = k
=k
2r
r
(2q)(2q)
q2
Ud = k
= 2k
2r
r
The electric potential energy of two point
charges approaches zero as the two point
charges move farther away from each other.
1. positive
If the three point charges shown here lie at
the vertices of an equilateral triangle, the
electric potential energy of the system of
three charges is
3. zero
2. negative
4. not enough information
given to decide
The electric potential energy of two point
charges approaches zero as the two point
charges move farther away from each other.
1. positive
If the three point charges shown here lie at
the vertices of an equilateral triangle, the
electric potential energy of the system of
three charges is
3. zero
If the length of the side of the triangle is d,
then
U = k ( q*q/d + q(-q)/d +q(-q)/d ) = - kq2/d < 0
2. negative
4. not enough information
given to decide
Which group of charges took more work to bring
together from a very large initial distance apart?
1
d
+2
3
+1
2
+1
d
+1
Both took the same amount of work
d
+1
d
Which group of charges took more work to bring
together from a very large initial distance apart?
1
2
d
+2
3
+1
+1
d
+1
Both took the same amount of work
The work needed to assemble
a collection of charges is the
same as the total PE of those
charges:
added over
all pairs
d
+1
d
For case 1:
only 1 pair
For case 2:
there are 3 pairs
These four 1.0 g spheres are released simultaneously and allowed to move away
from each other. What is the speed of each sphere when they are very far apart?
Ans (if needed, see next page): 0.49 m/s
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1. moves toward A with an increasing speed.
A proton is released from
rest at point B, where the
potential is 0 V.
Afterward, the proton
2. moves toward A with a steady speed.
3. remains at rest at B.
4. moves toward C with a steady speed.
5. moves toward C with an increasing speed.
1. moves toward A with an increasing speed.
A proton is released from
rest at point B, where the
potential is 0 V.
Afterward, the proton
2. moves toward A with a steady speed.
3. remains at rest at B.
4. moves toward C with a steady speed.
5. moves toward C with an increasing speed.
1. Va = Vb = Vc = Vd = Ve
Rank in order, from largest to
smallest, the potentials Va to
Ve at the points a to e.
2. Va = Vb > Vc > Vd = Ve
3. Vd = Ve > Vc > Va = Vb
4. Vb = Vc = Ve > Va = Vd
5. Va = Vb = Vd = Ve > Vc
1. Va = Vb = Vc = Vd = Ve
Rank in order, from largest to
smallest, the potentials Va to
Ve at the points a to e.
2. Va = Vb > Vc > Vd = Ve
3. Vd = Ve > Vc > Va = Vb
4. Vb = Vc = Ve > Va = Vd
5. Va = Vb = Vd = Ve > Vc
An AA battery is connected to a parallel plate capacitor having
4.0-cm-diameter plates spaced 2.0 mm apart. How much charge
does the battery supply to each plate?
Ans: 8.3 x 10-12 C
1) V > 0
What is the electric
potential at point A?
2) V = 0
3) V < 0
A
B
1) V > 0
What is the electric
potential at point A?
2) V = 0
3) V < 0
Since Q2 (which is positive) is closer
to point A than Q1 (which is negative)
and since the total potential is equal
to V1 + V2, then the total potential is
positive.
A
B
1) V > 0
What is the electric
potential at point B?
2) V = 0
3) V < 0
A
B
1) V > 0
What is the electric
potential at point B?
2) V = 0
3) V < 0
Since Q2 and Q1 are equidistant
from point B, and since they have
equal and opposite charges, then
the total potential is zero.
Follow-up: What is the
potential at the origin of the x-y
axes?
A
B
1. a point charge placed at P would feel no
electric force
Consider a point P in
space where the electric
potential is zero. Which
statement is correct?
2. the electric field at points around P is
directed toward P
3. the electric field at points around P is
directed away from P
4. none of the above
5. not enough information given to decide
1. a point charge placed at P would feel no
electric force
Consider a point P in
space where the electric
potential is zero. Which
statement is correct?
2. the electric field at points around P is
directed toward P
3. the electric field at points around P is
directed away from P
4. none of the above
5. not enough information given to decide
The value of the electric potential at any one
point in space is arbitrary. Only differences in
electric potential have physical significance.
The electric potential due to a point charge
approaches zero as you move farther away
from the charge.
1. positive
If the three point charges shown here lie at
the vertices of an equilateral triangle, the
electric potential at the center of the triangle
is
3. zero
2. negative
4. not enough information
given to decide
The electric potential due to a point charge
approaches zero as you move farther away
from the charge.
1. positive
If the three point charges shown here lie at
the vertices of an equilateral triangle, the
electric potential at the center of the triangle
is
3. zero
If the distance from any charge to
the center of the triangle is d, then
V = k ( q/d + q/d -q/d ) = kq/d > 0
2. negative
4. not enough information
given to decide
Four point charges are
arranged at the corners
of a square. Find the
electric field E and the
potential V at the center
of the square.
1) E = 0
V=0
2) E = 0
V≠0
3) E ≠ 0
V≠0
4) E ≠ 0
V=0
5) E = V regardless of the value
-Q
+Q
-Q
+Q
Four point charges are
arranged at the corners
of a square. Find the
electric field E and the
potential V at the center
of the square.
1) E = 0
V=0
2) E = 0
V≠0
3) E ≠ 0
V≠0
4) E ≠ 0
V=0
5) E = V regardless of the value
The potential is zero: the scalar
contributions from the two positive
charges cancel the two minus charges.
-Q
+Q
-Q
+Q
However, the contributions from the
electric field add up as vectors, and
they do not cancel (so it is non-zero).
Follow-up: What is the direction
of the electric field at the
center?
Example: electric potential of an electric dipole with dipole
moment p a distance r much larger than the separation 2a
bet ween the charges, r >> 2a.
kq k(−q)
V (P ) =
+
r1
r2
!
"
1
1
= kq
−
r1
r2
kq(r2 − r1 )
=
r1 r2
P
r2
r r
1
sθ
a
~2
co
–q
or, for a << r
k(2qa) cos θ
cos θ
V (r, θ) ≈
= kp 2
2
r
r
a
θ
a
+q
1
At which point
does V = 0?
5) all of them
2
+Q
3
4
–Q
1
At which point
does V = 0?
5) all of them
2
+Q
3
–Q
4
All of the points are equidistant from both charges. Since
the charges are equal and opposite, their contributions to
the potential cancel out everywhere along the mid-plane
between the charges.
Follow-up: What is the direction of the electric field at all 4
points?
Equipotential surfaces of a dipole
–
(show applet if possible)
+
Which of these configurations gives V = 0 at all
points on the x-axis?
+2µ
C
+1µ
C
+2µ
C
+1µ
C
x
-2µ
C
1)
-1µ
C
+2µ
C
-2µ
C
x
-1µ
C
4) all of the above
2)
-2µ
C
x
-1µ
C
5) none of the above
+1µC
3)
Which of these configurations gives V = 0 at all
points on the x-axis?
+2µ
C
+1µ
C
+2µ
C
+1µ
C
x
-2µ
C
1)
-1µ
C
+2µ
C
-2µ
C
x
-1µ
C
4) all of the above
2)
-2µ
C
x
-1µ
C
+1µC
3)
5) none of the above
Only in case (1), where opposite charges lie
directly across the x-axis from each other, do
the potentials from the two charges above the
x-axis cancel the ones below the x-axis.
Which of these configurations gives V = 0 at all
points on the y-axis?
+2µ
C
+1µ
C
+2µ
C
+1µ
C
x
-2µ
C
1)
-1µ
C
+2µ
C
-2µ
C
x
-1µ
C
4) all of the above
2)
-2µ
C
x
-1µ
C
5) none of the above
+1µC
3)
Which of these configurations gives V = 0 at all
points on the y-axis?
+2µ
C
+1µ
C
+2µ
C
+1µ
C
x
-2µ
C
1)
-1µ
C
+2µ
C
-2µ
C
x
-1µ
C
4) all of the above
2)
-2µ
C
x
-1µ
C
+1µC
3)
5) none of the above
Only in case (3), where opposite charges lie
directly across the y-axis from each other, do
the potentials from the two charges above the
y-axis cancel the ones below the y-axis.
Follow-up: Where is V = 0 for configuration #2?
1) A and C
Which two points
have the same
potential?
2) B and E
3) B and D
4) C and E
5) no pair
A
C
B
E
Q
D
1) A and C
Which two points
have the same
potential?
2) B and E
3) B and D
4) C and E
5) no pair
Since the potential of a point charge is:
A
C
only points that are at the same distance
from charge Q are at the same potential.
This is true for points C and E.
They lie on an Equipotential Surface.
Follow-up: Which point has the smallest
potential?
B
E
Q
D
Rank in order, from largest to
1. ∆V12 > ∆V13 = ∆V23
smallest, the potential differences
2. ∆V13 > ∆V12 > ∆V23
∆V12, ∆V13, and ∆V23 between
points 1 and 2, points 1 and 3, and
points 2 and 3. Here the symbol
∆V12 means ∆V12=V1-V2, etc
3. ∆V13 > ∆V23 > ∆V12
4. ∆V13 = ∆V23 > ∆V12
5. ∆V23 > ∆V12 > ∆V13
Rank in order, from largest to
1. ∆V12 > ∆V13 = ∆V23
smallest, the potential differences
2. ∆V13 > ∆V12 > ∆V23
∆V12, ∆V13, and ∆V23 between
points 1 and 2, points 1 and 3, and
points 2 and 3. Here the symbol
∆V12 means ∆V12=V1-V2, etc
1 and 2 the
same distance
from q: V1=V2
3. ∆V13 > ∆V23 > ∆V12
4. ∆V13 = ∆V23 > ∆V12
5. ∆V23 > ∆V12 > ∆V13
3 is farther away
since V = kq/r
potential at 3 is
smaller than at 1&2
Example: ring of charge Q , radius a,
with uniform charge density, on points along ring’s axis.
dq
Continuous charge
distribution:
x2 + a2
a
P
V =
x
(COB)
kQ
V (x) = √
a2 + x2
!
dV =
!
k dq
r
A thin rod with charge Q has been bent into a semicircle of radius
R. Find an expression for the electric potential at the center.
Ans: V = k Q/R
Example: disk of charge Q , radius a,
with uniform charge density, on points along disk’s axis.
Sum over concentric rings, using
previous result for ring of:
dr
r
P
x
radius:
r
charge:
dq = (2πr dr)σ = (2πr dr)
(COB)
V (x) =
!
2kQ "
a2
#
x2 + a2 − |x|
Q
Q 2
=
dr
2
2
πa
a
additional problems
At the back of a TV picture tube, a uniform electric field of 600 kN/C extends over
region of 5.0 cm and points towards the back of the tube. (a) Find the potential
difference bet ween the back and the front of this field region. (b) How much work
would it take to move an ion with charge +2e from the back to the front of the field
region? (c)What would happen to an electron released at the back of the field region
(COB)
Ans:
(a) 30 kV
(b) 9.6 x 10-15 J (or (2e)(30 kV) = 60 keV, see next page)
An isolated, infinite charged sheet carries a uniform charge density σ.
(a) Find an expression for the potential difference from the sheet to a
point a perpendicular distance x from the sheet.
(b) What is the potential difference bet ween t wo points the same
distance from the sheet?
(COB)
ANS:
∆V0x
σx
=−
2"0
A
E
E=
σ
2"0