Quest Electrostatics 1 Key

Version 001 – Electrostatics I – tubman – (12125) This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. AP EM 1993 MC 55 001 10.0 points Two metal spheres that are initially uncharged are mounted on insulating stands, as shown. −

− −−

X

Y

1

Two charged particles of equal magnitude (−Q and −Q) are fixed at opposite corners of a square that lies in a plane (see figure below). A test charge −q is placed at a third corner. −Q

−q −Q What is the direction of the force on the test charge due to the two other charges? 1.

A negatively charged rubber rod is brought close to but does not make contact with sphere X. Sphere Y is then brought close to X on the side opposite to the rubber rod. Y is allowed to touch X and then is removed some distance away. The rubber rod is then moved far away from X and Y. What are the final charges on the spheres? Sphere X

2. 3.

correct

4. 5.

Sphere Y 6.

1. Positive

Positive

2. Positive

Negative correct

7.

3. Negative

Negative

8.

4. Zero

Zero

5. Negative

Positive

Explanation: When the negatively charged rod moves close to the sphere X, the negatively charged electrons will be pushed to sphere Y. If X and Y are separated before the rod moves away, those charges will remain on X and Y, so X is positively charged and Y is negatively charged. AP EM 1998 MC 39 40 002 (part 1 of 2) 10.0 points

Explanation: The force between charges of the same sign is repulsive and between charges with opposite signs is attractive. −Q

−q

−Q

The resultant force is the sum of the two vectors in the figure.

Version 001 – Electrostatics I – tubman – (12125) 003 (part 2 of 2) 10.0 points If F is the magnitude of the force on the test charge due to only one of the other charges, what is the magnitude of the net force acting on the test charge due to both of these charges? √ 1. Fnet = 2 F correct 2. Fnet =

2F 3

3. Fnet = 0 F 4. Fnet = √ 2 3F 5. Fnet = 2

2

−8 µC placed on the x-axis at −2 cm ? The Coulomb constant is 8.9875 × 109 N · m2 /C2 . Correct answer: 139.806 N. Explanation: Let : q1 q2 q3 x1 x2 x3

= −9 µC = −9 × 10−6 C , = 8 µC = 8 × 10−6 C , = −8 µC = −8 × 10−6 C , = 4 cm = 0.04 m , = 10 cm = 0.1 m , and = −2 cm = −0.02 m .

Coulomb’s law (in vector form) for the electric force exerted by a charge q1 on a second charge q3 , is ~ 13 = ke q1 q3 ˆr13 , F r2 where ˆr13 is a unit vector directed from q1 to q3 ; i.e., ~r13 = ~r3 − ~r1 .

6. Fnet = 3 F 2F 7. Fnet = √ 3 8. Fnet = F 9. Fnet = 2 F F 10. Fnet = √ 3 Explanation: The individual forces form a right angle, so the magnitude of the net force is p √ Fnet = F 2 + F 2 = 2 F .

−10 −8 −6 −4 −2

2 4 x →

8 µC

−9 µC

−8 µC

Charge on Third Particle 004 10.0 points A particle with charge −9 µC is located on the x-axis at the point 4 cm , and a second particle with charge 8 µC is placed on the x-axis at 10 cm .

6 8 10 (cm)

What is the magnitude of the total electrostatic force on a third particle with charge

x13 = x3 − x1 = (−2 cm) − (4 cm) = −0.06 m x23 = x3 − x2 = (−2 cm) − (10 cm) = −0.12 m x3 − x1 x ˆ 13 = p = −ˆı (x3 − x1 )2 x3 − x2 x ˆ 23 = p = −ˆı (x3 − x2 )2

Since the forces are collinear, the force on the third particle is the algebraic sum of the forces between the first and third and the second and third particles: ~ =F ~ 13 + F ~ 23 F   q2 q1 = ke 2 ˆr13 + 2 ˆr23 q3 r13 r23 = 8.9875 × 109 N · m2 /C2  (−9 × 10−6 C) (−ˆı) × (−0.06 m)2  (8 × 10−6 C) + (−ˆı) (−0.12 m)2 ×(−8 × 10−6 C) = −139.806 N ,

Version 001 – Electrostatics I – tubman – (12125) with a magnitude of 139.806 N . Charges in a Thundercloud 005 (part 1 of 2) 10.0 points In a thundercloud there may be an electric charge of 40 C near the top of the cloud and −40 C near the bottom of the cloud. If these charges are separated by about 4 km, what is the magnitude of the electric force between these two sets of charges? The value of the electric force constant is 8.98755 × 109 N · m2 /C2 . Correct answer: 8.98755 × 105 N. Explanation:

3

The negative sign indicates that the force is attractive; i.e., the charges are being pulled towards one another. Compare Two Coulomb Forces 007 10.0 points Two charges q1 and q2 are separated by a distance d and exert a force F on each other. What is the new force F ′ , if charge 1 is increased to q1′ = 5 q1 , charge 2 is decreased q2 to q2′ = , and the distance is decreased to 2 d ′ d = ? 2 5 1. F ′ = F 2 2. F ′ = 50 F

Let :

q1 = 40 C , q2 = −40 C , d = 4 km = 4000 m , and k = 8.98755 × 109 N · m2 /C2 .

Considering the two sets of electric charges as point-charges, separated by a distance of 4 km, we can apply Coulomb’s law: q1 q2 F =k 2 r = 8.98755 × 109 N · m2 /C2 (40 C) (−40 C) × (4000 m)2 = −8.98755 × 105 N , with a magnitude of 8.98755 × 105 N . 006 (part 2 of 2) 10.0 points The electrostatic force between the top and the bottom of this thundercloud is

3. F ′ = 10 F correct 4. F ′ = 20 F 5. F ′ =

5 F 4

6. F ′ = 25 F 7. F ′ = 5 F 8. F ′ = 100 F 25 F 4 25 10. F ′ = F 2 Explanation: 9. F ′ =

k q1′ q2′ = F = r′ 2 ′

1. repulsive. = 10 2. undetermined. 3. zero. 4. attractive. correct Explanation:



k (5 q1 )  2 d 2

q2 2



k q1 q2 = 10 F . d2

Conceptual 16 Q03 008 10.0 points If you double the charge on one of two charged objects, how does the force between them change?

Version 001 – Electrostatics I – tubman – (12125) 1. Triples

2. Triples

2. Halves

3. No change correct

3. Does not change

4. Doubles

4. Doubles correct

5. Halves

4

Explanation: The charge on each object doubles, as does the distance:

5. Quadruples Explanation: q1 q2 ∝ q1 q2 r2 The factor is (2 q1 ) q2 = 2 (q1 q2 ) . The force doubles because the product of the charge doubles. Fe = ke

Conceptual 16 Q02 009 10.0 points If you double the distance between two charged objects, by what factor is the electric force affected?

F ∝

q1 q2 (2 q1 ) (2 q2 ) q1 q2 = = 2 . 2 2 r (2 r) r

The electrical force between them does not change. Induction and Grounding 011 10.0 points 1) Two uncharged metal balls, Y and X, each stand on a glass rod and are touching.

Correct answer: 0.25.

X

Y

Explanation: Fe = ke

1 q1 q2 ∝ 2 2 r r

The factor is 1 1 1 = = 0.25 2 . 2 2 (2 r) 4r r Conceptual 16 Q14 010 10.0 points Object A and object B are initially uncharged and are separated by a distance of 1 meter. Suppose 10,000 electrons are removed from object A and placed on object B, creating an attractive force between A and B. An additional 10,000 electrons are removed from A and placed on B and the objects are moved so that the distance between them increases to 2 meters. By what factor does the electric force between them change? 1. Quadruples

2) A third ball carrying a negative charge, is brought near the first two. −

X

Y

3) While the positions of these balls are fixed, ball Y is connected to ground. −

X

Y

4) Then the ground wire is disconnected. −

X

Y

Version 001 – Electrostatics I – tubman – (12125) 5) While Y and X remain in touch, the ball carring the negative charge is removed. X

Y

6) Then ball Y and X are separated. X

Y

5

charge (equally if identical). Finally we separate Y and X. The signs of the charge on Y and that on X are both positive. Force Between Electrons 012 10.0 points Two electrons in an atom are separated by 1.7 × 10−10 m, the typical size of an atom. What is the force between them? The Coulomb constant is 9 × 109 N · m2 /C2 . Correct answer: 7.99226 × 10−9 N.

After these procedures, what are the signs of the charge qY on Y and qX on X? 1.

qY is negative and qX is neutral

2.

qY is neutral and qX is negative

3.

qY is positive and qX is positive correct

4.

qY is positive and qX is neutral

5.

qY is negative and qX is positive

6.

qY is negative and qX is negative

7.

qY is positive and qX is negative

8.

qY is neutral and qX is positive

9.

qY is neutral and qX is neutral

Explanation: When the ball with negative charge is brought nearby, the free charges inside Y and X rearrange themselves. The positive charges are attracted and go to the right (i.e., move to X), leaving negative charges on the left hand side of the system Y Y, i.e., in Y. When we ground Y, electrons flow from the ground to Y (making it neutral), whereas the positive charges in X are still held enthralled by the negative charge on the third ball. We break the ground. Now we remove the third ball with negative charge. The charge on X is redistributed in the system Y Y; i.e., they share the positive

Explanation: Let :

d = 1.7 × 10−10 m and ke = 9 × 109 N · m2 /C2 .

The force between the electrons is ke q1 q2 ke qe2 = d2 d2 9 9 × 10 N · m2 /C2 = 1.7 × 10−10 m2 × (1.602 × 10−19 C)2

F =

= 7.99226 × 10−9 N .

Hewitt CP9 22 E20 013 10.0 points How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their original distance of separation? 1. doubles 2. Reduces to one quarter of original value 3. quadruples correct 4. Reduces to one half of original value 5. Doesn’t change Explanation: By the inverse square law, at half the distance the electric force field is four times as strong:

Version 001 – Electrostatics I – tubman – (12125) F ∝

1 1 2

r


2 =

4 r2

6