PHY 2005, Test 1, January 24, 2003

 

1. A charge of +3.0 microC is at x = 0, and a charge of -3.0 microC is at x = 6.0 m.  A third charge of -4.0 microC is placed at (x,y) = (3 m, 4 m).  What is the magnitude of the force on this third charge because of the other two charges?

 

(1) 0.73 x 10-3 N  (2) 1.5 x 10-3 N  (3) 4.1 x 10-3 N  (4) 5.2 x 10-3 N (5) 8.1 x 10-3 N

 

The solution is to compute the force acting on the -4.0 microC charge caused by each of the other two charges and to add the two forces using vector addition.  From the Pythagorean Theorem, the distance from the +3.0 microC to the -4.0 microC charge is 5 m, as is the distance between the -3.0 microC and the -4.0 microC charge.  Thus, the forces F1 and F2 are identical in magnitude, and the angle between F1 and Fnet is the same as the angle between F2 and Fnet.  From the symmetry and the geometry, you should see that the vertical component of the net force is zero, and the horizontal component is (2)(F1)cos(theta), where theta is the angle between F1 and Fnet.

Using Coulomb’s law and noting that cos(theta) = 3/5, the answer is

Fnet = (2)(9 x 109)(12/25)(10-12)(3/5)

 

2. Refer back to the previous question where a charge of +3.0 mC is at x = 0, and a charge of -3.0 mC is at x = 6.0 m.   What is the direction of the electric field at the point (x,y) = (3 m, 4 m) due to these two charges?

 

(1) in the +x direction  (2) in the –x direction  (3) in the +y direction  

(4) in the –y direction   (5) none of the choices

 

You can just notice that the direction of the electric field is the same as the direction of the force on a positive charge.  Since the force on a negative charge is to the left, the direction of the field must be to the right.

 

3. A positive charge q is placed a small distance d from a conducting sphere.  Which one or more of the following statements are true, depending on the sizes of the charges and the resulting amount of polarization?

a) If the sphere has a net positive charge, the charge and the sphere must repel each other.

b) If the sphere has no charge, the charge and the sphere must attract each other.

c) If the sphere has net negative charge, the charge and the sphere must attract each other.

 

(1) a   (2) b   (3) c   (4) a and b   (5) b and c

 

The reason a) is wrong is that the attraction due to polarization can be either stronger than the force of repulsion between the +q charge and the positively charged sphere, or it can be weaker, depending on the details. 

 

4. An alpha particle is composed o1f 2 protons and two neutrons (which have no charge).  The alpha particle is accelerated through a potential difference of 500 V.  How much kinetic energy is gained by the alpha particle?

 

(1) 500 eV   (2) 250 eV   (3) 1000 eV   (4) 2000 eV   (5) none of these

 

The gain in KE is equal to the loss in electrical potential energy.  This energy lost is qV, which is transformed into kinetic energy.  Thus, gain in KE = (2 electronic charges)(500 V) = 1000 eV

 

5. A dust particle having an unknown charge of q and a known mass of 2.0 x 10-8 kg is suspended against gravity between two very large charged conducting plates that are separated by 0.20 m.  The electrical potential difference between the plates is 80 volts.  What is q?

 

(1) 5 x 10-10 C  (2) 4 x 10-9 C  (3) 1 x 10-9 C  (4) 2 x 10-8 C  (5) 8 x 10-7 C

 

The downward gravity force, mg, is balanced by the upward electrical force, qE.  But the E-field is V/d = (80 V)/(0.20 m) = 400 V/m = 400 N/C.  Thus q = (2.0 x 10-8 kg)(10 m/s2)/(400 V/m) = 5 x 10-10 C.

 

6. Under equilibrium conditions, which one or more of the following are true for a solid metal sphere carrying a charge -Q?

a) The charge is uniformly distributed throughout the sphere.

b) The electric field inside the sphere is zero.

c) The electric field just outside the surface of the sphere is pointed radially outward.

 

(1) a  (2) b  (3) c  (4) a and b  (5) b and c

 

Your book says clearly that a) is wrong , since all the charges are on the surface).  Statement c) is wrong because the sphere is negatively charged, which means the field points radially inward.

 

7. Three capacitors CA = 2.0 mF, CB = 6.0 mF and CC = 3.0 mF are connected (in the given order) to each other in series.  A 12-V battery is connected across the system.  What is the charge on the plates of the 6.0-mF capacitor.

 

(1) 1.0 mC  (2) 2.0 mC  (3) 4.0 mC  (4) 6.0 mC  (5) none of these

 

First find the effective capacitance of the three series capacitors.  You should get Ceff = 1.0 microF.  A 12-V battery connected to Ceff results in a stored separated charge of Q = CV = 1.0 microC.  Each of the three capacitors in series must have this same charge on their plates, so Q = 12.0 microC.

 

8. The three capacitors are now placed in parallel.  The 12-V battery is connected across the system.  What is the potential difference across the 6.0-mF capacitor.

 

(1) 1 V  (2) 3 V  (3) 4 V  (4) 6 V  (5) none of these

 

All capacitors in parallel have the same voltage!  Thus, V = 12 V for each one.

 

9. A beam of electrons in a TV picture tube is measured to be 5.0 x 10-8 A.  How many electrons per second strike the screen from this beam?

 

(1) 8.0 x 109   (2) 1.6 x 1011  (3) 8.0 x 1011  (4) 3.1 x 1011  (5) 5.0 x 108

 

The ampere is charge per second, so that I = 5.0 x 10-8 C/s.  An electronic charge is 1.6 x 10-19 C.  Therefore, divide the electronic charge into the current to obtain the number of electrons/second.

 

10. Refer to the circuit shown.  Find the current through the battery.

 

(1) 0.25 A   (2) 0.33 A   (3) 0.42 A   (4) 0.67 A  (5) 0.75 A

 

The 4-ohm and 12-ohm resistors are in parallel.  Their effective resistance is 3 ohms.  The 3 ohms is in series with the 12-ohm resistor for a total of 15 ohms.  The current through the battery is

I = V/R = (10 V)/(15 ohms) = 2/3 A.

 

11. Refer to the same circuit.  Find the power expended in the 4-ohm resistor.

 

(1) 1.0 W   (2) 2.0 W   (3) 3.0 W   (4) 4.0 W   (5) none of these

 

The voltage across the 12-ohm resistor is IR = (2/3 A)(12 ohms) = 8 V.

The voltage across the 4- and 12-ohm resistors must then be 10 V - 8 V = 2 V.

You can also find the voltage across the parallel segment by using the effective resistance of the parallel part (IReff = (2/3 A)(3 ohms) = 2 V).

 

Since there are 2 V across the 4-ohm resistor, the power expended is P = V2/R, or

                P = (22)/4 = 1.0 watt

 

 

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