Chapter 21 online HW solutions and explanations posi tive charge Question 1. A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive density λ = 1 nC/m. nC/m. A second long, thin rod parallel to the z-axis is located at x = +1 cm and carries a uniform negative charge density λ = - 1 nC/m. What is the electric field at the origin? (1.8 × 103 N/C) 0 (3.6 × 103 N/C) ȋ (-1.8 × 103 N/C) (-3.6 × 103 N/C) Explanation: Start by drawing the rods as described in the question. question. You will notice the symmetry. One rod is placed at x = -1 cm while the other is placed at + 1 cm. Now remember that the electric field lines produced by such charged wires are perpendicular everywhere to the wires themselves. Further, the electric field arrows point away from the wire when charged positively and towards the wire when charged negat ively.
For this problem, at the origin, the positive rod produces an electric field line pointing away from the wire with a magnitude that can be calculated using the equation E = k
9
= (9x10 )
.
2(110 9 1.8 10 3 N / C (make sure everything is converted to SI units). Likewise, the E k 9 10 r 0.01 2
9
negative rod produces a similar electric field vector pointing towards it (because it’s negative). The magnitude of the second vector is the same as the first because we have the same linear charge density and the same 3 distance to the origin. Therefore, the two arrows add up to produce a net electric field Enet = 2E = 3.6 x 10 N/C along the positive x-direction.
would be the answer if the two wires were positively charges or Variations of the problem. Now think what would negatively charged. Perhaps, both parallel parallel to the y-axis y-axis and so on. Remember whatever you do, do, the rules described in the first paragraph still apply.
Question 2. The force of attraction between a -40.0 -40.0 μC and +108 μC charge is 4.00 N. What is the separation between these two charges? 2.49 m
3.12 m 2.10 m 3.67 m 1.13 m Explanation: -6 -6 Here, Q1 = 40 x 10 C; Q2 = 108 x 10 C and F = 4N
Then applying coulomb’s law: F k
Q1 Q2 r 2
40 10 6 108 10 6 4 9 10 2 r
9
2
2
We realize that the only unknown is the distance r that I am looking for. This leads to r = 9.709m or r = 3.12 m
Variations of the problem: There’s not much to vary. We just need to understand that we’re dealing with coulomb’s law and that it’s not always the force that we’re after. In this example, we gave you the distance between the two charges as an unknown variable but the charges themselves could become unknown in other problems.
Question 3. A proton is placed in an electric field of intensity 700 N/C. What is the magnitude and direction of the acceleration of this proton due to this field? 6.71 × 1010 m/s2 opposite to the electric field
67.1 × 1010 m/s2 in the direction of the electric field 6.71 × 1010 m/s2 in the direction of the electric field 67.1 × 1010 m/s2 opposite to the electric field 6.71 × 109 m/s2 opposite to the electric field Explanation: First thing to remember is that a positive charge w ould move in the same direction as the electric field it’s subjected to. A negative charge would move in the opposite direction to the field.
In the electric field, the proton feels the electric force and gravity. The proton is very tiny so the force of gravity is negligible as it is the case with the vast majority of such problems. Therefore, the net force acting on the charge is the electric force F = qE. This same force causes the proton to accelerate and move in the same direction as the electric field (remember it’s a positive charge). Using Newton’s second law, we have F = ma. This means that qE = ma. The only unknown in this equation is the acceleration that I am looking for. -19
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10
2
a = qE/m p = 1.6 x 10 x 700/(1.67 x 10 ) =6.71 x 10 m/s in the direction of the electric field. Variations of the problem. What if you change the proton into an electron? What will happen? Of course the mass will be different but also the direction of travel would be opposite the direction of the electric field. How about neutrons?
Question 4. An electric dipole of dipole moment P 5 ×10 -10 C.m i is placed in an electric ˆ
field E 2 ×10 6 N/mi 2 ×10 6 N/m j . What is the maximum torque experienced by the dipole?
ˆ
ˆ
1.40 × 10-3 N.m 1.00 × 10-3 N.m 0 N.m 2.80 × 10-3 N.m 2.00 × 10-3 N.m Explanation:
We know the simple torque equation: p E . Whatever method you use, you should be able to calculate this cross product which would result in:
p E 5 ×10 -10 C.m i 2 ×106 N/m i 2 ×10 6 N/m j ] 110 3 N.m
ˆ
ˆ
ˆ
Question 5. Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E midway between these two charges? 25.2 × 105 N/C directed towards the positive charge
25.2 × 106 N/C directed towards the positive charge 25.2 × 104 N/C directed towards the negative charge 25.2 × 106 N/C directed towards the negative charge 25.2 × 105 N/C directed towards the negative charge Explanation: This is very easy by now. We did many similar examples in class. The first thing to do is plot the charges. Then, locate the midpoint. Label the distances to each charge which would 0.1 m to each charge. Next draw the electric field contributions one at a time. Start for example with the positive charge. A vector E+ at midpoint should point away from the charge. A vector E – starting at the midpoint location should point toward the negative charge. So we have two electric field contributions that point in the same direction. These will add up. -6
2
7
E+ = k (20 x 10 )/(0.1) = 1.798 x 10 N/C towards negative charge (or away from the positive charge) -6 2 6 E – = k (8 x 10 )/(0.1) = 7.198 x 10 N/C towards negative charge Thus the total electric field is E T = E+ + E – = 25.2 × 106 N/C directed towards the negative charge. Variations of the problem. Try making these charges both positive or both negative. Perhaps you want to calculate the electric field at a point other than the middle, etc Question 6. Three point charges are located at the following positions: Q1 = 2.00 μC at x = 1.00 m; Q2 = 3.00 μC at x = 0; Q3 = -5.00 μC at -1.00 m. What is the magnitude of the force on the 3.00 μC charge? 0.189 N
8.10 × 10-2 N 5.40 × 10-2 N 0.158 N 0.135 N Explanation: Again this is very simple. Plot the charges. We have a linear 1D problem with three charges one after the other. I am interested in the net force on the middle positive charge. I know that this charge will be feeling the other two charges so there will be two forces acting on it. Let’s start pairing it with the left charge which is also positive, so we would expect repulsive forces. It’s very important that you know how to draw these forces. One force (vector) F12 is on charge Q1 pointing left and another equivalent force vector F21 is drawn on Q2 (the middle charge) pointing right. The magnitudes of F12 and F21 are the same. Remember Newton’s third law that forces happen in pairs due to the principle of action and reaction.
Now pair Q2 with the charge on the right side Q3. This last one is negative so we have a pair of attractive forces F23 and F32 (again same magnitude because of the action-reaction principle). Draw the forces properly making sure that one of these F23 is located on Q2 and pointing right (towards Q3). If you drew everything properly you will see that F21 and F23 point in the same direction so they must add.
9
-6
-6
2
F21 = (9x10 ) (|2x10 | x |3x10 |)/(1 ) and 9 -6 -6 F23 = (9x10 ) (|3x10 | x |5x10 |)/(12) 9
-12
Thus, F = F21 + F23 = (9x10 ) x (21 x 10 ) = 0.189 N Variations of the problem: Change the sign of Q3 to positive. Play with the signs and try to calculate the net force on the other charges as well. Just remember that every one of these three charges will feel a total of 2 forces because it sees two other charges. Question 7. Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 8.0 cm, what is the electric force between them? 16 N
1.0 N 4.0 N 1/16 N 1/4 N Explanation: Here I have no clue what the magnitudes of the two charges are! However, I have two cases to compare. I also should know that Coulomb’s law remain v alid (whether I know the numerical values or not). Let’s write Coulomb’s law for the first case: Q1 Q2 1 9 10 9 2 2 r 0.02 Now for the second case: F 1 k
F 2 k
Q1 Q2
Q1 Q2 r 2
Q1 Q2 F 2 9 10 9 0.082
If you look hard at these two equations, you will realize that Q1 and Q2 are the same. They can be dropped by taking the ratio of the two equations (in other terms divide F2 by F1. 2
2
Thus F2/F1 = r 1 /r 2 or F2 = 1/16 N Variations of the problem: You may try other numbers if you wish but you will realize that the last expression will always be true. Question 8. A metal sphere of radius 2.0 cm carries a charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sphere? 5.7 × 106 N/C
3.4 × 106 N/C 7.5 × 106 N/C 4.2 × 106 N/C 9.3 × 106 N/C Explanation: This is a metal sphere. The electric field inside is zero. However, we’re asked to determine the field OUTSIDE the sphere (d = 6 cm > than the radius of the sphere R = 2 cm).
Now remember that a metal (conducting) sphere or even a non-conducting sphere both behave the same on the outside. Both will act as point charges as if all the charge were at the center of the sphere. This means that my distances must be measured from the center of the sphere (not from the surface). 2
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2
The Electric Field equatin is, thus, E = kq/r = k |3 x 10 |/(0.06) = 7.5 × 106 N/C Variations of the problem: What is the electric field at the surface of the sphere? What is the electric field inside the sphere (see above)? What is the electric field at some distances (choose reasonable values) inside and outside a non-conducting sphere? Question 9. Two point charges of magnitudes +5.00 μC, and +7.00 μC are placed along the x-axis at x = 0 cm and x = 100 cm, respectively. Where must a third charge be placed along the x-axis so that it does not experience any net force because of the other two charges? 50 cm
45.8 cm 91.2 cm 9.12 cm 4.58 cm Explanation: At least in this problem, the author made my life easy. I only need to check locations on the x-axis. Draw the two charges as described above. You will see you only have three options, left of the origin, between the two charges, and to the right of the second charge (i.e., beyond 1 m).
We don’t know what is the sign of the third unknown charge so we have to experiment with both positive and negative. Let’s take a positive charge and place to the left of the origin. You will immediately realize that it will experience two repulsive forces due to the other two positive charges. These two forces will always (on that side of the x-axis) be in the same direction and cannot cancel each other. Now place the same third charge to the right of the second charge. You will realize that we have the same situation. We are only left with one option: between the two charges. Place it there and you will see that the two repulsive forces acting on the third charge point opposite each other. These could eventually be cancelled by sliding the third charge somewhere in between. The condition that needs to be met is F31 = F32 (in magnitude). Remember how to read these subscripts. F31 means the force on charge#3 due to charge #1 and so on. Thus, at some point r between the two charges, a third charge will experience net force equal to zero when F31 = F32 2
2
Or, kqQ1/r = kqQ2/(100-r)
2
2
(100-r) /r = 7/5
r = 45.8 cm
Variations of the problem. Try a negative charge instead. Try changing the sign of one or the other of the two known charges. Draw the force diagrams (and solve if you wish).
Question 10. Two point charges each have a value of 3.0 C and are separated by a distance of 4.0 m. What is the electric field at a point midway between the two charges?
9.0 × 107 N/C 3.6 × 107 N/C Zero 18 × 107 N/C 4.5 × 107 N/C Explanation: We did something similar above. However, now we have a very special case. The charges are identical and the distance from the point of interest (midpoint) to either charge is the same so we would expect the electric field contributions to be exactly the same. Since both charges are positive the two electric field vectors (one from each charge) point in opposite direction. Therefore, they cancel out. Variations of the problem: try making these charges negative! Or maybe one positive and the other negative (that would be similar to another problem we solved above). Question 11. A 10 microcoulomb charge is placed at the origin and a 20 microcoulomb charge is placed on the x-axis at x = 40 cm. If an electron is placed on the y-axis at y = 30 cm, what is the magnitude of the force it will experience? 4.92 × 10-13 N
0.615 × 10-13 N 2.46 × 10-13 N 2.46 × 10-14 N 1.23 × 10-13 N Explanation: This is a 2D problem. Draw the charges and the forces acting on the electron. You should end up with two forces on the electron. One force vector pointing in the negative y-direction (down towards the positive charge at the origin) and the other force pointing towards the other positive charge (that would be along the hypotenuse of the right triangle formed by the three charges).
To add these two vectors you will need to find their components and follow the vector addition rules. -6 -19 2 -6 -19 2 F31 = k |10x10 ||1.6x10 |/(0.3) , F32 = k |20x10 ||1.6x10 |/(0.5) Variations of the problem: Try calculating the net force on each of the other charges.
Question 12. A charge Q = 3 μC is located at the origin. The electric field created by this charge at a point of coordinates (x = 2 m, y = 0 m) is equal to ________. (0.75 × 103 N/C) j
(13.5 × 109 N/C) i (6.75 × 103 N/C) i (6.75 × 103 N/C) j (13.5 × 103 N/C) j Explanation:
This is a positive charge. At the requested location, the y-coordinate is zero. Therefore, I am really looking at a point charge on the x-axis and nothing else matters. The electric field produced at that point should point away from the charge so it will be in the positive x-direction. To calculate the magnitude, we just use the simple point-like electric field equation 2 -6 3 E = kQ/r = k 3x10 /4 = 6.75 x 10 N/C in the positive x-direction (or ȋ ) Variations of the problem: What if y #0? In that case, you would need to calculate the actual distance from the charge to that point and use in the electric field equation. You would get the magnitude but not the direction. To get the direction, find the x and y components of the electric field and use tan()=Ey/Ex to find the angle with respect to the x-axis. Question 13. An atomic nucleus has a charge of +40e. An electron is 10-9 m from the nucleus. What is the force on the electron? 9.2 nN
1000 C 2.9 nN 6.8 nN 3.7 nN Explanation: Couldn’t be any simpler. Two charges interacting with each other. Use Coulomb’s law. 2 9 -19 -19 -9 2 -9 F = k |q||Q|/r = (9x10 ) |40x1.6x10 ||1.6x10 |/ (1x10 ) = 9.2 x 10 N Question 14. An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q = 1 μC, is located at the point (x, y, z) = (0, 1 cm, 0) while the negative charge is located at the point (x, y, z) = (0, -1 cm, 0). How much work will be done by an electric field E = (3 × 106 N/C ) i to bring the dipole to its stable equilibrium position? 0.02 J
0J 0.06 J 0.12 J 0.03 J Explanation: First locate the two charges on the xy axes. You will realize that the dipole is along the y-axis and that the dipole moment p points upwards. The electric field is in the positive x-direction (notice the ȋ ). This means that the angle between p and E is 90 degrees.
The work done by the external electric field to rotate the dipole and bring into stable equilibrium equals the change in the dipole energy or U = Uf - Ui. Now remember that the energy of the dipole U = -pEcos. We have the magnitude of E and we know o o the starting angle = 90 and the final angle = 0 . We don’t have the dipole moment magnitude p but -6 -8 we can calculate it knowing that p = qd = 10 (C) x (0.02m) = 2x10 C m. o
o
-8
6
Thus the work done W = U = -pE (cos90 – cos0 ) = (2x10 ) x (3x10 ) = 0.06 J Variations of the problem: try other angles or starting and ending points.
Question 15. A dipole moment is placed in a uniform electric field oriented along an unknown direction. The maximum torque applied to the dipole is equal to 0.1 N.m. When the dipole reaches equilibrium its potential energy is equal to -0.2 J. What was the initial angle between the direction of the dipole moment and the direction of the electric field? 0
90° 10° 45° 30° Explanation: Here, the problem is giving me information when the dipole is at two distinct states. One when the dipole is oriented at some unknown direction with respect to the electric field so we consider the angle between the dipole moment and E to be simply , and second we know the dipole’s energy when it’s in o stable equilibrium which means that the dipole and the electric field are fully aligned with = 0 . The magnitudes of p and the electric field E are both unknown. However, since we have two situations to consider, it’s likely that I could use the result of o ne equation to substitute into the second.
Let’s apply the torque and dipole energy equations: pE sin and U pE cos . The maximum torque that the dipole experiences in this problem does not mean the absolute maximum that could be o achieved if the dipole were allowed to go to other positions (i.e. when = 90 ). Now plug things in:
0.1 pE sin and 0.2 pE cos 0 pE 0.2
Plug in the first equation and solve for the angle: q = sin
1 0.1
30 0.2
Question 16. An electric field is set up between two parallel plates, each of area 2.0 m2, by putting 1.0 μC charge on one plate and a -1.0 μC charge on the other. The plates are separated by 4.0 mm. What is the magnitude of the electric field between the plates at a distance of 1.0 mm from the positive plate? 5.6 × 104 N/C
0 N/C 1.4 × 104 N/C 3.1 × 104 N/C 4.2 × 104 N/C Explanation: First, we should remember that the electric f ield inside such a device that we called “capacitor” is constant. Outside, it’s zero. The constant electric field inside (anywhere inside) can be calculated using this equation: E = σ/ϵo Thanks god I am not left guessing what σ (surface charge density) I should use. I know that all the charges will be on the inner sides of the two plates (when facing each other – please check my slides). All I need to do is calculate σ myself since I know the total charge on each plate and surface area of each plate. σ = Q/A = 0.5 x 10-6 C/m2 and E = σ/ϵo = (0.5 x 10-6) / (8.85x10-12) = 5.6 x 104 N/C
Variations of the problem: what would the electric field magnitude be 1 mm from the negative plate inside the capacitor? How about the center? outside?
Question 17. A metal sphere of radius 10 cm carries a charge of +2.0 μC. What is the magnitude of the electric field 5.0 cm outside the sphere's surface? 8.0 × 107 N/C
4.0 × 107 N/C 8.0 × 105 N/C 4.2 × 106 N/C 4.0 × 105 N/C Explanation: We did something similar above. It’s important to remember that I need to measure the distance from the center of the sphere to the point of interest. That distance is 10 cm + 5 cm = 15 cm or 0.15 m. Now, outside the sphere, this thing acts as a point -charge. 2
9
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2
5
Use the corresponding electric field equation: E = k|Q|/r = (9x10 ) |2x10 | / (0.15) = 8.0 x 10 N/C Question 18. Consider a square which is 1.0 m on a side. Charges are placed at the corners of the square as follows: +4.0 μC at (0, 0); +4.0 μC at (1, 1); +3.0 μC at (1, 0); -3.0 μC at (0, 1). What is the magnitude of the electric field at the square's center? 1.3 × 105 N/C
1.5 × 105 N/C 1.9 × 105 N/C 1.1 × 105 N/C 1.7 × 105 N/C Explanation: Before doing anything, please examine the distances and charge magnitudes and then plot the electric field contributions at the center. You should have 4 vectors at the center. The ones obtained with the positive 4 C charges should have the same lengths while the other two obtained with the positive or negative 3 C charges should be shorter. Nonetheless, these last two shorter vectors are also equal in length.
Now see how the symmetry helps you cancel the field vectors due to charges at (0,0) and (1,1). All I need to worry about is the other pair wher clearly the two vectors are pointing in the same direction towards the negative charge. 1/2
The distance from any charge to the center is: r = [(1+1) ]/2 = 0.71 m The net electric field at the center due to the charges at (1,0) and (0,1) is: 2 9 -6 2 5 E = 2 x (kq/r ) = 2 (9x10 ) (3x10 ) / (0.71) = 1.1 x 10 N/C. The factor 2 in the equation above came from the fact that both vectors have the same length so we just multiplied the magnitude of one vector by 2 to get the total. Variations of the problem: there’s a huge potential for variations here. You could easily make all charges the same and positive or negative. You could also create pairs of different signs and so on.
Question 19. Three equal charges are located in the x-y plane with coordinates (0, 3m), (4m, 3m), and (4m, 0). The direction of the force on the charge located at (4m, 3m) is the same as the direction of ________
4 i + 3 j. 9 i + 16 j. 16 i + 9 j. 12 i + 6 j. 3 i + 4 j. Explanation: Plot the charges on the xy axes. Let’s label the charges as 1, 2, and 3, respectively. The one we’re interested in is the middle one so it’s charge #2. This problem could’ve been much harder. However, the choice of the charge at (4,3) makes things relatively simple and easy. We know by now that this charge will be subjected to two forces due to the other charges, namely F21 and F23. F21 is a force vector on charge 2 pointing left towards charge 1. Likewise, F23 is a force vector on the same charge 2 pointing downward towards charge 3. Notice that these two forces are completely within the x and y axes so they simply make the x and y components of the net force on charge 2. 2
2
2
2
2
2
Fnet = F21 + F23 = kq [r13/r 13 + r23/r 23 ] = kq [i /4 + j /3 ] = (kq2/144) (9i + 16 j) unit vectors.)
(here r13 and r23 are
Notice in the above equation the bold font used for the forces which means we’re dealing with vectors. Variations of the problem: calculate the net force on each of the remaining charges. Those will be a bit trickier but we’ve done similar problems in class. Question 20. An atomic nucleus has a charge of +40e. What is the magnitude of the electric field at a distance of 1.0 m from the nucleus? 6.0 × 10-8 N/C
5.8 × 10-8 N/C 6.2 × 10-8 N/C 5.4 × 10-8 N/C 5.6 × 10-8 N/C Explanation: Very simple! We have a point like charge and I need to calculate the electric field at a distance of 1 m away from it. Use the appropriate point-charge equation: 2 9 -19 -8 E = k q/r = (9x10 ) x |40 x1.6x10 |/ 1 = 5.8x10 N/C Variations of the problem: What is the direction of the electric field at that point. Would the magnitude of the electric field increase or decrease as I move away the charge? Toward the charge?
Question 21. A thin, circular disk of radius R = 30 cm is oriented in the yz-plane with its center as the origin. The disk carries a total charge Q = +3 μC distributed uniformly over its surface. Calculate the magnitude of the electric field due to the disk at the point x = 15 cm along the x-axis.
4.98 × 105 N/C 3.32 × 105 N/C 2.49 × 105 N/C 9.95 × 105 N/C 1.99 × 105 N/C Explanation: Plot the problem. Identify the electric field direction. This is a very simple problem if I know the
appropriate equation to calculate the electric field on the axis of a disk. For such a disk, we know that
E x
2
x 1 where σ is the surface charge density, x is the distance from the center of the disk 1 2 2 2 x R )
to any point on its axis, and R is the radius of the disk. 2
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2
2
First calculate σ = Q/(πR ) = (3 x 10 C)/[(0.3) ]. Remember that (πR ) is the surface area of a disk. Then calculate E x= 3.32 × 105 N/C
Variations of the problem: what is the direction of the electric field at that point? What if we did the
same calculation at a point on the axis below the disk? What happens if I change the sign of the disk charge? How about bringing 2 disks and placing them parallel to each and try to calculate the field at the midpoint? Question 22. If the earth's electric field is 100 N/C downward, what must be the charge on a 1 kg object so that it would be "weightless"? 0.22 C
1.0 C 9.8 × 10-2 C -9.8 × 10-2 C 9.8 C Explanation: First draw a sketch showing the earth and a small object above the surface. Draw the field lines pointing downward as described above. What does the earth look like? (a negative charge of course).
Now think of the forces acting on the object. We have gravity and the electric field. For this object to be weightless (to float) the net force should be zero. We know that the force of gravity always point towards the ground so we conclude that the electric force on the charged object must be pointing upward (in the opposite direction to gravity). Since the electric field must be upward (opposite the direction of the electric field as well) we conclude that the charge on it must be negative (so I should look for a negative answer). For weightlessness, the two opposite forces (electric and gravity) must be equal: Thus, mg = qE
-2
Solving for the magnitude of q: q = mg/E = (1)(9.8)/100 = 9.8 x 10 C but it’s negative so it must be (-2 9.8 x 10 C). Question 23. A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? 0
2.50 × 1013 1.25 × 1013 2.50 × 1019 1.25 × 1019 Explanation: For the plastic object to have a net positive charge, we must have removed electrons from its atoms. This leaves me with excess protons compared to the electron count.
To quantify how many more protons we have, we use the charge quantization equation q = Ne with q being the total net charge, N the number of the charge carriers (protons here) and e the elemental charge -19 carried by the proton or (1.6 x 10 C). -6
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13
N = q/e = (2 x 10 ) / (1.6 x 10 ) = 1.25 x 10 protons. Question 24. Two charges Q1 = -1 μC and Q2 = +1 μC are located on the y-axis at y1 = +1 cm and y2 = -1 cm respectively. A third charge Q3 = +8 μC is added on the y-axis so that the electric field at the origin is equal to zero. What is the position of Q 3? y3 = +8 cm
y3 = -2 cm y3 = +2 cm y3 = +4 cm y3 = 0 cm
Explanation: We did something above kind of similar in concept but with electric forces. Remember that electric forces are not the same as electric fields. Make sure you know how to separate the concepts and how to plot and calculate each. Let’s start with the first 2 charges and figure what kind of electric field they produce at the origin. Charge one is negative and it is located on the positive side of the y-axis. At the origin, it would produce an electric field vector E1 pointing upward (towards it because it ’s negative). Now, charge 2 is positive and located on the negative side of the y axis. At the origin, it would produce and electric field vector E2 also pointing upward (away from it because it’s positive). It’s clear that that the two vectors E1 and E2 add in the upward direction. Let’s calculate those. 9
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2
7
E1 = (9x10 )|1x10 |/(1x10 m) = 9x10 N/C E2 must have the same magnitude since we have a similar charge magnitude (forget about the sign) and the same distance to the origin.
E1+E2 = 18x107 N/C Now a third positive charge must be placed on the y axis in such a way it produces a strong enough electr ic field at the origin to cancel E1+E2. We obviously need a vector E3 pointing downward with the same magnitude as E1+E2 or 18x107 N/C. Since the charge is positive, the only way to produce a vector pointing downward at the 2
origin is to place the third charge above the origin. Now using the equation E = k |q|/r 7
9
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2
We should be able to calculate the distance r on the y-axis as follows: E3 = (18x10 ) = (9x10 )|8x10 |/r . 2
9
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7
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2
Or r = (9x10 )|8x10 |/(18x10 ) = 4 x 10 m which gives r = 0.02 m or 2 cm.
Variations of the problem: What of the third charge were negative? Where should it be located?
