Download DP Physics Topic 5 Quiz #1 Name: Show formulas, substitutions

DP Physics Topic 5 Quiz #1
Show formulas, substitutions, and units!
Name: _____________________________
A balloon becomes charged to 125 C by rubbing it on Albert the Physics Cat.
1. What is the sign of the charge the balloon acquires?
negative
2. How many electrons are transferred between the balloon and Albert?
125 x 10-6/1.6 x 10-19 = 7.8 x 1014 electrons
3. How do the charges of the balloon and Albert compare?
Same quantity opposite sign
The following questions are about electric current.
4. A +125 C charge is moved past a point in a conductor in 25.0 ms. What is the value of the electric
current involved in this movement?
I = q/t = 125 x 10-6 / 25 x 10-3 = 5 x 10-3 amps = 5mA
5. An electrical discharge between a cloud and a lightning rod has a current of 1250 A for a time of 1.25
ms. How much electric charge was involved in this lightning strike?
q = It = 1250* 1.25 x 10-3= 1.56 C
Two equal point charges of +125 C are placed 0.625 mm apart.
6. If the charges are located in air, or vacuum, what is their electric force? Is it attractive or repulsive?
Fe = (8.99 x 109)(125 x 10-6)2 / (0.625 x 10-3)2 = 3.60 x 108 N (repulsive – same charge)
7. If the charges now have a 0.625 mm layer of mica between them, what is the new electric force
between them? Assume this purity of mica has a permittivity of 7.50 times that of free space.
Fe = 1/(4 * 7.5 * 8.85 x 10-12) (125 x 10-6)2 / (0.625x 10-3 )2 = 4.80 x 107 N
A conducting sphere of radius 0.125 m holds an electric
charge of Q = +80.0 C. A charge q = +20.0 C is located
in the vicinity of Q.
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8. Find the electric force between the two charges if
q is located 0.25 m from the surface of Q.
F = (8.99 x10 9)(80.0 x 10-6)(20.0 x 10-6)/(.125 + .25)2 = 102 N
9. Find the electric force between the two charges if q is moved onto the surface of Q.
F = (8.99 x10 9)(80.0 x 10-6)(20.0 x 10-6)/(.125)2 = 921 N
10. If the mass of q is 0. 250 g what is its initial acceleration if released from this new position?
(F = ma)
F = ma  a = F/m = 921 / .00025 = 3.68 x 106 m/s2
A +50 C charge is located in free space.
11. Find the electric field strength 0.25 m from the charge.
E = kQ/r2 = (8.99 x 109)(50 x 10-6)/.252 = 7.2 x 106 N
12. Find the force acting on an electron  that is 0.25 m from the charge. Is it attractive, or is it
repulsive?
F = (8.99 x 109) (50 x 10-6)(1.6 x 10-19)/.252 = 1.0 x 10-12 attractive
Lab Activities
13. Explain why the aluminum foils in the electroscrope repel each other even if the charged object did
not touch the spiral part of the copper wire.
Through induction, the electrons at the top are pushed down into the aluminum foils and sign
liked charges repel, the aluminum foils are pushed apart.
14. Explain the relationship between the equipotential lines and the electric field lines. (hint: from last
week lab)
The equipotential lines are perpendicular to the electric field lines.