1. Two charges are placed at opposite ends of a meter stick. Approximately where could a third charge be placed so it would be in equilibrium?
2. A 2 C charge and a 8 C charge repel each other with 10 N of force. How much will a 4 C charge and a 4 C charge repel each other when placed the same distance apart?
A. 40 N B. 20 N C. 10 N D. 8 N
5. Consider a 75 W light bulb connected to a 120 V outlet.
a. Find the resistance of the light bulb when lit.
b. The tungsten filament (wire) in the light bulb has a diameter of 1.8x10-5 m2 and a resistivity of 8.2x10-8 W.m at the operating temperature. Find the length of the filament.
6. Find the equivalent resistance of the circuit shown below. (R1 is on the top left, R3 and R4 are on the right side of the square on the same line and R2 is down the middle line.)
R2 =14 W
R1 = 12 W
R4 = 10 W
R3 = 16 W
9. Nichrome wire of cross-sectional radius 0.791 mm is to be used in winding a heating coil. If the coil must carry a current of 9.25 A when voltage of 1.20 x 10^2 V is applied across its ends, find (a) the required resistance of the coil and (b) the length of wire you must use to wind the coil.
10. Suppose your waffle iron is rated at 1.00 kW when connected to a 1.20 x 10^2 V source. (a) What current does the waffle iron carry? (b) What is its resistance?
11. What minimum of 75-W light bulbs must be connected in parallel to a single 120-V household circuit to trip a 30.0-A circuit breaker?
12. Three 9.0 Ώ resistors are connected in series with a 12-V battery. Find (a) the equivalent resistance of the circuit and (b) the current in each resistor. (c) Repeat for the case in which all three resistors are connected in parallel across the battery.
Please refer to the attachment.
1. Two charges are placed at opposite ends of a meter stick as shown in the fig.. Approximately where could a third charge be placed so it would be in equilibrium?
q1= -2.5μC q2 = -8.5μC
Solution: Let the third charge +Q be placed at a distance x from q1 for it to be at equilibrium.
Then from Coulomb's law magnitude of force on Q due to q1 is given by:
F1 = kq1Q/x2 where k = 1/4Πε0
F1 = k(-2.5x10-6)Q/x2
Taking force acting towards right as +ve and that towards left as -ve, the direction of F1 is negative. Hence,
F1 = - k(-2.5x10-6)Q/x2 ..........(1)
Force on Q due to q2 = F2 = kq2Q/(1 - x)2 = k(-8.5x10-6)Q/(1 - x)2
As F2 acts towards right, it is taken as +ve.
For equilibrium, vector sum of F1 and F2 must be zero. Hence,
- k(-2.5x10-6)Q/x2 + k(-8.5x10-6)Q/(1 - x)2 = 0
+2.5/x2 - 8.5/(1 - x)2 = 0
2.5(1 - x)2 - 8.5x2 = 0
1 + x2 - 2x - 3.4x2 = 0
2.4x2 + 2x - 1 = 0
Solving for x: x = [-2 + √22 - 4(2.4)(-1)]/2(2.4) = [-2 + √4 + 9.6]/4.8
x = (-2 + 3.69)/4.8
Taking + sign: x = 0.352 m
Taking - sign: x = - 1.18 m
As Q must lie between the two charges, x must be +ve, x = 0.352 m is the correct solution.
2. A 2 C charge and a 8 C charge repel each other with 10 N of force. How much will a 4 ...
This solution provides assistance with the physics problems attached.