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Version 047 – Midterm1-v1 – li – (55755) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A neutral copper ball is suspended by a string. A positively charged insulating rod is placed near the ball, which is observed to be attracted to the rod. Why is this? 1. There is a rearrangement of the electrons in the ball. correct 2. The number of electrons in the ball is greater than in the rod. 3. The ball becomes positively charged by induction.

1

8.98755 × 109 N · m2/C2 . 1. 38743.1 2. 810892.0 3. 166484.0 4. 1196650.0 5. 76144.5 6. 402509.0 7. 347769.0 8. 621318.0 9. 844516.0 10. 212034.0 Correct answer: 4.02509 × 105 N/C. Explanation:

Let : q = −42.8 µC = −4.28 × 10−5 C , r = 78 cm = 0.78 m , and k = 8.98755 × 109 N · m2 /C2 . y

4. The string is not a perfect conductor.

−−− A −− θ −− r − E − − O − − − −− −−− B

5. The ball becomes negatively charged by induction. Explanation: If a positively charged insulator is brought close to a conductor, it will attract some of the free electrons of the conductor closer to the insulator. The attraction between the insulator and the conductor results from the rearrangement of electrons. 002 10.0 points Consider a charged semicircular arc with radius 78 cm and total charge −42.8 µC distributed uniformly on the semicircle. y

y

∆θ

−−− A −− θ −− r − − − O − − − −− −−− B

II

I

III

IV

x

x

Find the magnitude of the electric field at O . The value of the Coulomb constant is

y

∆θ

II

I x

III

IV

x

By symmetry of the semicircle, the y component of the electric field at the center is Ey = 0 . We need consider only the x-component of the electric field, so ds = r dθ q q ∆q = λ ds = λ r dθ = r dθ = dθ π πr k |∆q| cos θ k |q| cos θ ∆θ , ∆Ex = = π r2 r2 and the magnitude of the electric field at the center is Z π/2 2 k |q| k |q| cos θdθ = E = Ex = 2 π r2 −π/2 π r =

2 (8.98755 × 109 N · m2 /C2) π (0.78 m)2 × |(−4.28 × 10−5 C)|

= 4.02509 × 105 N/C

Version 047 – Midterm1-v1 – li – (55755) with direction along the negative x axis. 003 10.0 points A 108 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 4.72 × 105 N · m2 /C. What is the electric field strength? 1. 3889410.0 2. 1877530.0 3. 515234.0 4. 338748.0 5. 679017.0 6. 438540.0 7. 1611850.0 8. 1470720.0 9. 4929830.0 10. 2652580.0 Correct answer: 5.15234 × 105 N/C.

Explanation:

Let : r = 54 cm = 0.54 m and Φ = 4.72 × 105 N · m2/C . I ~ · dA ~ . The position of By Gauss’ law, Φ = E

maximum electric flux will be that position in which the plane of the loop is perpendicular ~ · dA ~ = E dA. to the electric field; i.e., when E Since the field is constant,

If before the sphere C is introduced the electrostatic force between A and B is Fi , how does Ff , the electrostatic force between A and B after C is removed, compare to Fi ? 1. Ff = 0 Fi 1 Fi 8 1 3. Ff = Fi 16 1 4. Ff = Fi 4

2. Ff =

5. Ff = Fi 3 Fi 4 3 7. Ff = Fi correct 8 1 8. Ff = Fi 2 3 Fi 9. Ff = 16 Explanation: Since the two conducting spheres are identical (i.e., same radius), when the spheres touch the charges redistribute themselves equally between the two spheres. Let spheres A and B have an initial charge Q. When an identical uncharged sphere C comes in contact with sphere A and is removed, then by conservation of charge, each sphere will carry charge 6. Ff =

QC1 = QA =

2

Φ = EA= E πr 4.72 × 105 N · m2 /C Φ E= 2 = πr π (0.54 m)2

1 Q. 2

When sphere C touches sphere B, then each sphere will carry charge

= 5.15234 × 105 N/C . 004 10.0 points Two identical conducting spheres, A and B , carry equal charge. They are stationary and are separated by a distance much larger than their diameters. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and finally removed (to a far away distance).

2

QC2

1 Q+Q QC1 + QB 3 = QB = = 2 = Q. 2 4 2

Hence if the initial force is given by Fi = ke then the final force is

Q2 , d2

Version 047 – Midterm1-v1 – li – (55755)

Ff = ke



  1 3Q Q 3 2 4 = Fi . 2 8 d

005 10.0 points What can we conclude from observing an attractive force between a positively charged rod and some ob ject?

1. The object has a net negative charge. 2. Cannot be determined correct 3. The object is an insulator. 4. The object has a net positive charge. 5. The object is a conductor. Explanation: Repulsion would guarantee the object has a net positive charge, but attraction occurs if the object has a net negative charge, or no net charge. Both insulators and conductors will be attracted to a charged rod, as we saw in class.

3

2qsQ z2 qsQ 3. − k 2 z qsQ 4. + k 2 z qsQ 5. + k 3 z 2qsQ correct 6. − k z3 2qsQ 7. + k z2 qsQ 8. − k 3 z Explanation: Using the expression for the field along the dipole axis and noting that the force must be repulsive since the negative dipole charge is closest, we have   +k2qs 2qsQ F = −Q × = −k . 3 z z3 2. − k

007

10.0 points

006 10.0 points Consider the setup shown in the figure.

Determine the force the dipole exerts on -Q. Choose the correct expression for the force. + sign is along the positive x-direction. 1. + k

2qsQ z3

A solid plastic ball has negative charge uniformly spread over its surface. Which of the diagrams in the above figure best shows the polarization of molecules inside the ball? 1. (d)

Version 047 – Midterm1-v1 – li – (55755) 2. (c) 3. (a) correct

4

3. sphere 1 accelerates 16 times as fast as sphere 2. 4. sphere 1 accelerates 4 times as fast as sphere 2.

4. (b) Explanation: Firstly, let us note that plastic serves as an insulator. Next, we note that the electron clouds of neutral atoms are repelled by the negative charge on the surface. Hence, of the diagrams shown in the picture, (a) represents the best possible choice. 008 10.0 points A charge of +1 Coulomb is place at the 0cm mark of a meter stick. A charge of −1 Coulomb is placed at the 100-cm mark of the same meter stick. Is it possible to place a proton somewhere on the meter stick so that the net force on it due to the two charges is 0? 1. Yes; to the right of the 50-cm mark 2. No correct 3. Yes; to the left of the 50-cm mark Explanation: The proton is repelled from the +1 Coulomb charge and attracted to the −1 Coulomb charge. Both forces act in the same direction so they cannot cancel out. 009 10.0 points Two spheres, fastened to “pucks”, are riding on a frictionless airtrack. Sphere 1 is charged with 1 nC, and sphere 2 is charged with 4 nC. Both objects have the same mass. 1 nC is equal to 1 × 10−9 C. As they repel, 1. sphere 2 accelerates 16 times as fast as sphere 1. 2. sphere 2 accelerates 4 times as fast as sphere 1.

5. they have the same magnitude of acceleration. correct 6. they do not accelerate at all, but rather separate at constant velocity. Explanation: The force of repulsion exerted on each mass is determined by F =

1 Q1 Q2 = ma 4 π ǫ0 r2

where r is the distance between the centers of the two spheres. Since both spheres have the same mass and are subject to the same force, they have the same acceleration. 010 10.0 points Four point charges, each of magnitude 15.32 µC, are placed at the corners of a square 97.1 cm on a side. If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. The value of Coulomb’s constant is 8.98755 × 109 N · m2/C2 . 1. 7.20183 2. 0.102026 3. 0.163057 4. 7.97861 5. 6.40031 6. 120.368 7. 104.438 8. 13.3741 9. 2.11848 10. 4.28263 Correct answer: 4.28263 N. Explanation: Let : ke = 8.98755 × 109 N · m2/C2 , d = 97.1 cm = 0.971 m , and Q = 15.32 µC = 1.532 × 10−5 C .

Version 047 – Midterm1-v1 – li – (55755) 1 +

2 + F14

+ 3 The forces are

F34

F24 4

Q2 F34 = F24 = ke 2 d   = 8.98755 × 109 N · m2 /C2  2 1.532 × 10−5 C × 2 (0.971 m) = 2.23728 N and ke Q2 1 1 ke Q2 F14 = √ 2 = = F34 . 2 2 2 d 2d

The vector sum of F24 and F34 is in the same direction as F14 , so q F = F14 + F342 + F24 2   q 1 1 √ 2 + 2 F34 = F34 + 2 F34 = 2 2   1 √ + 2 (2.23728 N) = 2 = 4.28263 N . 011 10.0 points A negative point charge −Q is at the center of a hollow insulating spherical shell, which has an inner radius R1 and an outer radius R2. There is a total charge of +3 Q spread uniformly through out the volume of the insulating shell, not just on its surface. Determine the electric field at r, where r > R2 . kQ k3Q + r2 R22 kQ k3Q 2. E = + 2 − R1 R22 k2Q 3. E = R22 k2Q 4. E = R12

5

kQ k3Q + R21 R22 k3Q 6. E = 2 r kQ k3Q 7. E = − 2 + 2 r R1 k2Q 8. E = 2 correct r Explanation: According to Gauss’s Law, E 4 π r2 = Qinside , and Qinside = −Q + 3 Q = 2 Q. So ǫ0 the electric field is 2Q 2kQ . E= = 2 r2 4 πǫ0 r 5. E = −

012 (part 1 of 2) 10.0 points

A plastic hollow sphere, which is uniformly charged with negative charge on its surface, is placed near the center of a horizontal plastic rod, which is uniformly charged with negative charge. What is the direction of the electric field vector at P due to the charges on the plastic sphere alone?

1. E = −

1. direction 4 2. Zero magnitude correct

Version 047 – Midterm1-v1 – li – (55755)

6

7. direction 8 3. direction 2 8. direction 2 4. direction 3 9. direction 7 5. direction 1 6. direction 5 7. direction 6 8. direction 8 9. direction 7 Explanation: Since the charge distribution on the surface is uniform, the field at P due to the charges on the sphere alone is 0. 013 (part 2 of 2) 10.0 points What is direction of the resultant field vector contributed by the rod plus the sphere?

Explanation: The E field due to the charged shell is 0 and the E field due to the rod points downward. By the principle of superposition the correct answer is that the E-field points in direction 5. 014 10.0 points A uniformly charged circular arc AB of radius R is shown in the figure. It covers a quarter of a circle and it is located in the second quadrant. The total charge on the arc is Q > 0 and ∆s ≡ R ∆θ . y y

∆θ

++ A ++ + θ ++ R + + B O

II

I

III

IV

x x

What is the direction of the electric field ~ at the origin due to the charge disvector E tribution? 1. In quadrant IV correct 2. In quadrant III 3. In quadrant I 1. direction 3 2. direction 1 3. direction 5 correct 4. direction 4 5. direction 6 6. Zero magnitude

4. Along the negative y-axis 5. Along the positive y-axis 6. Along the positive x-axis 7. Along the negative x-axis 8. In quadrant II Explanation: The electric field for a positive charge is

Version 047 – Midterm1-v1 – li – (55755) directed away from it. In this case, the electric field generated by each ∆q will be directed into quadrant IV, so the total electric field will be in that quadrant. 015 10.0 points Two large, parallel, insulating plates are charged uniformly with the same positive areal charge density +σ, which is the charge per unit area. What is the magnitude of the resultant electric field E? The permittivity of free space 1 ǫ0 = . 4 π ke σ 1. Zero between the plates, outside corǫ0 rect σ σ outside between the plates, 2 ǫ0 2 ǫ0 σ 3. Zero between the plates, outside 2 ǫ0 2σ outside 4. Zero between the plates, ǫ0 σ 5. between the plates, zero outside 2 ǫ0 σ 6. everywhere ǫ0 σ between the plates, zero outside 7. ǫ0 2σ 2σ outside 8. between the plates, ǫ0 ǫ0 2σ 9. between the plates, zero outside ǫ0 2.

10. Zero everywhere Explanation: Each plate produces a constant electric field σ of E = directed away from the plate for 2 ǫ0 positive charge density, and toward the plate for negative charge density. Between the two plates, the two fields cancel each other so that Enet = 0. Outside the two plates, the fields add together, so that Enet =

016

σ . ǫ0

10.0 points

7

A positive charge q and a dipole of moment ~p contribute to a net field at location A that is zero, as shown in the following figure. ~p is a distance √ a from point A, while q is a distance b = 2a from A. Assume a ≫ dipole separation. b A

+ q

a

~p If ~p points to the right, we denote it +p; if ~p points to the left, we denote it −p. Choose the relationship between p and q that results in a net field of 0 at A. 1. p =

qa 2

2. p = qa 3. p = −

qa correct 2

4. p = −qa

Explanation: At A, the field due to the point charge points to the left. In order to get cancellation of E at A, the field due to the dipole must point to the right. Recall that, on the axis perpendicular to the dipole moment, the dipole field points in the direction opposite the dipole moment, so the dipole moment must point to the left: ~p = −p. To determine the magnitude of the relationship between p and q, we simply set the point charge field at A and the dipole field at A equal to each other: |Edip⊥ | = |Eq | kq k|~p| = √ 2 3 a ( 2a) qa . |~p| = 2 Combining the two results, our answer must qa be: p = − . 2

Version 047 – Midterm1-v1 – li – (55755)

8

so Gauss’s law gives 017 10.0 points A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radius r with r < R is used to calculate the magnitude of the electric field E at a distance r from the center of the sphere. Which equation results from a correct application of Gauss’s law for this situation? Q 1. E (4 π r2) = ǫ0   4 3 Q r2 2. E πr = ǫ0 R2 3   Q r 4 3 πr = 3. E 3 ǫ0 R2   4 3 Q r3 4. E πr = 3 ǫ0 R3 Q r3 5. E (4 π R2) = ǫ0 R3 Q 3 r3 6. E (4 π R2) = ǫ0 4 π R Q 7. E (4 π R2) = ǫ0 Q 3 r3 8. E (4 π r2) = ǫ0 4 π R 9. E (4 π r2) = 0 10. E (4 π r2) =

Q r3 correct ǫ0 R3

Explanation: Applying Gauss’s law, I I Qinside ~ E · ~n dA = E dA = . ǫ0 S S 4π 3 V = r ∝ r3 ; because charge Q is uni3 formly distributed, the enclosed charges are related by Qsurf ace r3 = 3 R Qsphere Qsurf ace =

r3 Qsphere , R3

E (4 π r2) =

Q r3 . ǫ0 R3...