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week 4
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![Page 3: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/3.jpg)
![Page 4: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/4.jpg)
![Page 5: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/5.jpg)
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
P1
2
3
4
![Page 6: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/6.jpg)
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.
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
P1
2
3
4
![Page 7: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/7.jpg)
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
P1
2
3
4
![Page 8: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/8.jpg)
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).
1) P → 1
2) P → 2
3) P → 3
4) P → 4
5) all require the same amount of work
P1
2
3
4
Follow-up: Which path requires the least work?
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.
![Page 9: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/9.jpg)
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
![Page 10: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/10.jpg)
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
![Page 11: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/11.jpg)
When a positive charge moves in the direction of the electric field,
1. the field does positive work on it and the potential energy increases
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
![Page 12: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/12.jpg)
When a positive charge moves in the direction of the electric field,
1. the field does positive work on it and the potential energy increases
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 >0The change in potential energy is the negtive of the work done by the field, so
ΔU = - W <0
![Page 13: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/13.jpg)
When a negative charge moves in the direction of the electric field,
1. the field does positive work on it and the potential energy increases
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
![Page 14: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/14.jpg)
When a negative charge moves in the direction of the electric field,
1. the field does positive work on it and the potential energy increases
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 <0The change in potential energy is the negtive of the work done by the field, so
ΔU = - W >0
![Page 15: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/15.jpg)
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
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?
electron
proton
electron
proton+
-
![Page 16: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/16.jpg)
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.
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?
electron
proton
electron
proton+
-
![Page 17: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/17.jpg)
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
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?
electron
proton
electron
proton+
-
![Page 18: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/18.jpg)
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.
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?
electron
proton
electron
proton+
-
![Page 19: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/19.jpg)
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
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?
electron
proton
electron
proton+
-
![Page 20: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/20.jpg)
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.
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?
electron
proton
electron
proton+
-
![Page 21: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/21.jpg)
Here is a simple problem we already know how to solve using forces and F=ma.But let’s solve it using conservation 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 105 m/s
![Page 22: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/22.jpg)
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.
• 1. Ua = Ub = Uc = Ud
• 2. Ud = Ub > Uc = Ua
• 3. Ud > Ub = Uc > Ua
• 4. Ub > Ua > Ud > Uc
• 5. Ud > Ub = Uc > Ua
![Page 23: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/23.jpg)
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.
• 1. Ua = Ub = Uc = Ud
• 2. Ud = Ub > Uc = Ua
• 3. Ud > Ub = Uc > Ua
• 4. Ub > Ua > Ud > Uc
• 5. Ud > Ub = Uc > Ua
Ua = kq2
rUb = k
2q2
rUc = k
(2q)q2r
= kq2
rUd = k
(2q)(2q)2r
= 2kq2
r
![Page 24: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/24.jpg)
The electric potential energy of two point charges approaches zero as the two point charges move farther away from each other.
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
1. positive
2. negative
3. zero
4. not enough information given to decide
![Page 25: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/25.jpg)
The electric potential energy of two point charges approaches zero as the two point charges move farther away from each other.
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
1. positive
2. negative
3. zero
4. not enough information given to decide
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
![Page 26: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/26.jpg)
Which group of charges took more work to bring together from a very large initial distance apart?
1
+1
+1
+1
d d
d
2+1+2
d
3 Both took the same amount of work
![Page 27: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/27.jpg)
The work needed to assemble a collection of charges is the same as the total PE of those charges:
Which group of charges took more work to bring together from a very large initial distance apart?
1
+1
+1
+1
d d
d
2+1+2
d
3 Both took the same amount of work
For case 1: only 1 pair
For case 2: there are 3 pairsadded over
all pairs
![Page 28: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/28.jpg)
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
![Page 29: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/29.jpg)
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?
1 2
3 4
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!"!!!E!<9!"!!9@!"!!@;!"!!;<F!"!E!<@!"!!9;F!
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Final energy is pure kinetic energy:
!"#$%#! &'()*+! "#$%&'!()!*+#)$%,$-.!/0$!*01%&$-!)20$%$)!1%$!2+(#3!*01%&$).!
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!"!!!E!<9!"!!9@!"!!@;!"!!;<F!"!E!<@!"!!9;F!
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![Page 30: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/30.jpg)
A proton is released from
rest at point B, where the
potential is 0 V.
Afterward, the proton
1. moves toward A with an increasing speed.
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.
![Page 31: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/31.jpg)
A proton is released from
rest at point B, where the
potential is 0 V.
Afterward, the proton
1. moves toward A with an increasing speed.
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.
![Page 32: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/32.jpg)
Rank in order, from largest to
smallest, the potentials Va to
Ve at the points a to e.
1. Va = Vb = Vc = Vd = Ve
2. Va = Vb > Vc > Vd = Ve
3. Vd = Ve > Vc > Va = Vb
4. Vb = Vc = Ve > Va = Vd
5. Va = Vb = Vd = Ve > Vc
![Page 33: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/33.jpg)
Rank in order, from largest to
smallest, the potentials Va to
Ve at the points a to e.
1. Va = Vb = Vc = Vd = Ve
2. Va = Vb > Vc > Vd = Ve
3. Vd = Ve > Vc > Va = Vb
4. Vb = Vc = Ve > Va = Vd
5. Va = Vb = Vd = Ve > Vc
![Page 34: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/34.jpg)
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
![Page 35: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/35.jpg)
1) V > 0
2) V = 0
3) V < 0
What is the electric potential at point A?
A B
![Page 36: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/36.jpg)
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.
1) V > 0
2) V = 0
3) V < 0
What is the electric potential at point A?
A B
![Page 37: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/37.jpg)
1) V > 0
2) V = 0
3) V < 0
What is the electric potential at point B?
A B
![Page 38: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/38.jpg)
Since Q2 and Q1 are equidistant
from point B, and since they have equal and opposite charges, then the total potential is zero.
1) V > 0
2) V = 0
3) V < 0
What is the electric potential at point B?
Follow-up: What is the potential at the origin of the x-y axes?
A B
![Page 39: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/39.jpg)
Consider a point P in
space where the electric
potential is zero. Which
statement is correct?
1. a point charge placed at P would feel no electric force
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
![Page 40: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/40.jpg)
Consider a point P in
space where the electric
potential is zero. Which
statement is correct?
1. a point charge placed at P would feel no electric force
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.
![Page 41: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/41.jpg)
The electric potential due to a point charge approaches zero as you move farther away from the charge.
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
1. positive
2. negative
3. zero
4. not enough information given to decide
![Page 42: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/42.jpg)
The electric potential due to a point charge approaches zero as you move farther away from the charge.
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
1. positive
2. negative
3. zero
4. not enough information given to decide
If the distance from any charge tothe center of the triangle is d, then
V = k ( q/d + q/d -q/d ) = kq/d > 0
![Page 43: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/43.jpg)
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
![Page 44: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/44.jpg)
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
The potential is zero: the scalar contributions from the two positive charges cancel the two minus charges.
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?
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.
![Page 45: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/45.jpg)
Example: electric potential of an electric dipole with dipole moment p a distance r much larger than the separation 2a between the charges, r >> 2a.
r2
!
P
r1r
–q +qa a
~2a co
s !
V (P ) =kq
r1+
k(!q)r2
= kq
!1r1! 1
r2
"
=kq(r2 ! r1)
r1r2
or, for a << r
V (r, !) ! k(2qa) cos !
r2= kp
cos !
r2
![Page 46: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/46.jpg)
At which point
does V = 0?
1
3
2
4
+Q –Q
5) all of them
![Page 47: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/47.jpg)
At which point
does V = 0?
1
3
2
4
+Q –Q
5) all of them
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?
![Page 48: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/48.jpg)
Equipotential surfaces of a dipole
+–
(show applet if possible)
![Page 49: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/49.jpg)
4) all of the above 5) none of the above
1)
x
+2µC
-2µC
+1µC
-1µC 2)
x
+2µC
-1µC
+1µC
-2µC 3)
x
+2µC
-1µC
-2µC
+1µC
Which of these configurations gives V = 0 at all points on the x-axis?
![Page 50: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/50.jpg)
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 x-axis?
4) all of the above 5) none of the above
1)
x
+2µC
-2µC
+1µC
-1µC 2)
x
+2µC
-1µC
+1µC
-2µC 3)
x
+2µC
-1µC
-2µC
+1µC
![Page 51: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/51.jpg)
Which of these configurations gives V = 0 at all points on the y-axis?
4) all of the above 5) none of the above
1)
x
+2µC
-2µC
+1µC
-1µC 2)
x
+2µC
-1µC
+1µC
-2µC 3)
x
+2µC
-1µC
-2µC
+1µC
![Page 52: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/52.jpg)
Which of these configurations gives V = 0 at all points on the y-axis?
4) all of the above 5) none of the above
1)
x
+2µC
-2µC
+1µC
-1µC 2)
x
+2µC
-1µC
+1µC
-2µC 3)
x
+2µC
-1µC
-2µC
+1µC
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?
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Which two points have the same potential?
1) A and C
2) B and E
3) B and D
4) C and E
5) no pair
A
C
B DEQ
![Page 54: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/54.jpg)
Which two points have the same potential?
1) A and C
2) B and E
3) B and D
4) C and E
5) no pair
A
C
B DEQ
Since the potential of a point charge is:
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?
![Page 55: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/55.jpg)
Rank in order, from largest to
smallest, the potential differences
∆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. ∆V12 > ∆V13 = ∆V23
2. ∆V13 > ∆V12 > ∆V23
3. ∆V13 > ∆V23 > ∆V12
4. ∆V13 = ∆V23 > ∆V12
5. ∆V23 > ∆V12 > ∆V13
![Page 56: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/56.jpg)
Rank in order, from largest to
smallest, the potential differences
∆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. ∆V12 > ∆V13 = ∆V23
2. ∆V13 > ∆V12 > ∆V23
3. ∆V13 > ∆V23 > ∆V12
4. ∆V13 = ∆V23 > ∆V12
5. ∆V23 > ∆V12 > ∆V13
1 and 2 the same distance from q: V1=V2
3 is farther away
since V = kq/r
potential at 3 issmaller than at 1&2
![Page 57: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/57.jpg)
Example: ring of charge Q , radius a, with uniform charge density, on points along ring’s axis.
dq
x
x2 + a2
aP V =
!dV =
!k dq
r
Continuous charge distribution:
(COB)
V (x) =kQ!
a2 + x2
![Page 58: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/58.jpg)
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
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Example: disk of charge Q , radius a, with uniform charge density, on points along disk’s axis.
xPr
dr
Sum over concentric rings, usingprevious result for ring of:
radius:
charge:
r
dq = (2!r dr)" = (2!r dr)Q
!a2=
Q
a2dr2
(COB)
V (x) =2kQ
a2
!"x2 + a2 ! |x|
#
![Page 60: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/60.jpg)
additional problems
![Page 61: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/61.jpg)
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 between 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)
![Page 62: week 4 - Physics Courses...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](https://reader034.vdocuments.us/reader034/viewer/2022043003/5f81cb7986f53d729e057204/html5/thumbnails/62.jpg)
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 between two points the same distance from the sheet?
(COB)
ANS:
!V0x = ! !x
2"0
EA
E =!
2"0