, where the sum is over all pairs of charges (q 1, q 2 ) that are


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1 February 12, 2010 PHY2054 Solutions Exam I 13. Particle 1 has a mass of 1.0 g and charge 6.0 µc, and particle 2 has a mass of 2.0 g and charge ( ) µc. While particle 2 is held in place, particle 1 is released from rest at a distance of 2.0 mm from particle 2. After a long time, what is the speed of particle 1 [in m/s]? Answer: Energy conservation determines that E f = ½ mv 2 = E i = kq 1 q 2 /d. Solve for v to get the answers Four point charges are placed at the four corners of a 3.0 cm 3.0 cm square, in two different configurations as shown in the figure. The magnitude Q of the charges is ( ) µc. Which configuration has higher electrical potential energy and what is its value [in J]? Answer: Configuration 1 ( ) The potential energy located a distance d apart., where the sum is over all pairs of charges (q 1, q 2 ) that are You can tell the answer for the highest energy configuration is 1 since here the like charges are closer by and like charges always make positive contribution to the potential energy, increasing it. The terms involving near pairs cancel in the U sum. Only the diagonals survive and lead to the answer Three point charges (Q 1 = 5.0 µc, Q 2 = 4.0 µc, and Q 3 = 6.0 µc) are located at the corners of an equilateral triangle, as shown. The electric potential due to these charges is ( ) V at the center of the triangle. If Q 2 is increased to 8.0 µc and Q 3 is decreased to 4.0 µc, what will be the electric potential at the center [in V]? Answer: ( ) The potential is given by the sum over all the charges which happen to be located at the same distance from the center of the triangle. Thus the sum is over only the charges (alternatively, the value given for one set of charges determines the size of the triangle). Therefore V 2 being the potential with new charges and V 1 the potential for original charges, In the electric circuit shown, capacitors 14 have the same capacitance of 2.0 µf each, and the amount of charge stored in capacitor 2 is 3.0 µc. What is the total energy stored in the four capacitors [in µj]?
2 Answer: ( ) Since E = ½ CV 2, we need the equivalent capacitance as well as the potential difference. The equivalent capacitance across the series part C = [1/2 +1/2+1/2] 1 = 2/3 µf, which in parallel with 2 µf gives C = 8/3 µf. The charge across capacitance 2 is 3µC. It must be the same charge across the net series capacitance, C = 2/3 µf and therefore the potential difference across the series link is Q/C = 9/2 = 4.5V. The energy stored in all capacitors is ½ CV 2 = ½ 8/ = 27 µj In the electric circuit shown, C 1 = 4.0 µf, C 2 = 8.0 µf, and C 3 = ( ) µf. The amount of charge stored in capacitor C x is exactly half the amount of charge stored in C 2. How large is C x [in µf]? Answer: ( ) The charge is the same in series. Thus the charge on C 2 is the same as the one on the series combination at left and likewise, the charge on C x is the same as the one on the series combination at right. Equivalent capacitances on each side should be in the ratio of the charges on C x and C 2. Which can be solved for C x = 2.4µF In the circuit shown, all resistance values are in ohms. Find the equivalent resistance between points a and b [in Ω]. Answer: (9 6 12) A matter of redrawing the circuits You bought two light bulbs, one rated ( ) W/120 V and the other 100 W/120 V. Then, you connected them in series and plugged the twobulb circuit into a 120 V power outlet as shown. Approximately, how much power is going to be dissipated by the circuit? Answer: ( ) W Let the power ratings of the two light bulbs be P 1 and P 2, and their resistances be R 1 and R 2. Then, P 1 =V 2 /R 1 and P 2 =V 2 /R 2, where V is the outlet voltage 120 V. When the light bulbs are connected in series as shown in figure, the power dissipated by each light bulb will be less than
3 the rated value because the potential drop across it will be less than V. (Equivalently, the decrease in the dissipated power occurs because the current through either of the two light bulbs decreases due to the resistance of the other light bulb in series.) The power dissipated by the two light bulbs is P=V 2 /(R 1 +R 2 ) because their total resistance is R 1 +R 2. This gives P=V 2 /(V 2 /P 1 +V 2 /P 2 )=1/(1/P 1 +1/P 2 ) An electric engineer designs a capacitor that would have two parallel plates of 1 cm 2 area each separated by a 100 µm thick dielectric ceramic sheet with a dielectric constant κ= (5 4 2) and resistivity ρ = ( ) Ωm. If such a capacitor is charged to 1 µc, estimate how long it will take for a half of the charge to leak through its slightly conductive ceramic sheet? Answer: (1 year 1 month 1 day) The capacitance of the capacitor is C=κε 0 A/d, where A is the area of the plate and d the distance between the two plates. At the same time, the resistance of the ceramic sheet is R=ρd/A. Therefore, the time constant of the charge decrease, τ = RC, is ρκε 0. The time it takes for the charge to decrease to ½ of its initial value is (ln2) τ, which in this case is (ln2) ρκε 0. The unit is sec. (1day = 86,400 sec) You have a piece of wire with the total resistance R. You interconnect the two wire ends to make a square loop and measure the resistance between points A and B as shown in the figure. Approximately, what is this resistance in terms of R? Answer: (0.2R 0.25R) In Problem 28, the resistance of the straight section between points A and B is R/4, whereas the resistance of the Cshaped section between the two points is 3R/4. These two resistances in parallel give the net resistance between the two points: 3R/16. Similarly, the two parallel resistances in Problem 29 are both R/2, resulting in the net resistance of R/ You have a piece of wire with the total resistance R. You interconnect the two wire ends to make an equilateral triangular loop and measure the resistance between points A and B as shown in the figure. Approximately, what is this resistance in terms of R? Answer: 0.22R In this problem, the two parallel resistances are R/3 and 2R/3, resulting in the net resistance of 2R/ Three identical charges (q = µc) lie along a circle of radius 2 m at angles of 40 o, 140 o and 270 o (as usual with respect to the x axis). What is the resultant electric field (in kilonewton/coulomb) at the center of the circle? Answer: ( ) ŷ
4 Let the radius of the circle be r. At the center of the circle, the net electric field produced by the charges at 40 o and 140 o is 2sin(40 o )k E q/r 2 in the negative y direction, since the x components of the individual fields due to the two charges cancel out exactly. The field produced by the charge at 270 o is k E q/r 2 in the positive y direction. The total field is therefore [2sin(40 o )1]k E q/r 2 in the negative y direction A proton (+e) shoots through tiny holes in two parallel metal plates across which ( ) V of electric potential difference is applied, as shown in the figure. The speed of the proton is originally v 0 = m/s as it goes through the hole in the first plate. Find its speed as it goes through the hole in the second plate. Answer: ( ) 10 5 m/s Let the applied electrical potential difference be V. Then the potential energy of the proton, as it goes through the hole in the second plate, is higher by e( V) than its potential energy when it went through the hole in the first plate. Therefore, the kinetic energy of the particle, as it goes through the hole in the second plate, is mv 2 /2 =mv 0 2 /2 e( V). Solve this for v Three equal charges each of magnitude (7 5 3) µc are arranged at the corners of an equilateral triangle, with sides of length 10 cm. Calculate the magnitude of the force acting on each of these charges. Answer: ( ) N In this geometry, the magnitudes of the forces acting on the three charges are the same. For convenience, let us position the triangle so that one charge is on the y axis and the two other charges are on the x axis, and consider the force on the first charge. This force is 2sin(60 o )]k E q 2 /r 2 in the y direction, where q is the magnitude of each charge and r the length of the sides of the triangle The electric field required to suspend a proton electron He ++ ion against the force of gravity is: Answer: 102 nanovolts/m directed upward 56 picovolts/m directed downward 205 nanovolts/m directed upward Since charge of the electron is negative, the electric field has to point down in order to exert an upward force against the force of gravity. For other particles, the direction of the field has to be upward. Let the mass of the particle be m and the magnitude of its charge q. The force of gravity on the particle is mg and the electrical force qe, where E is the magnitude of the electric field. Therefore, E=mg/q Three identical (size and shape) conductive balls, labeled A, B and C, carry charge 2q 2q q, q q 2q and 2q q 2q respectively. They are placed equidistant from each other. First A is brought to touch B and then separated back to the original position.
5 Next A and C touch each other and are restored to the old locations. Finally, B touches C and then returns to its place. What is the magnitude of the final force between A and B in proportion to the original force (F f =F i ) between them? Answer: 5/64 7/64 3/64 As an example, let us consider the first version of this problem, in which the initial charges on balls A, B, and C are 2q, q, and 2q. In the first step, in which balls A and B touch each other and separate, the total charge 3q will be equally divided, resulting in 3q/2 on each ball. In the second step, the total charge will be 3q/2 2q= q/2, which will be equally divided, resulting in q/4 on each ball. In the last step, the total charge 3q/2 q/4 = 5q/4 will be equally divided into two parts, 5q/8 each. After these three steps, the charges on A and B will be respectively q/4 and 5q/8. Originally, the force between A and B was k E (2q)q/r 2, where r is the distance between the two balls. After the redistribution of the charges, the force between A and B will be k E ( q/4) (5q/8)/r 2, whose magnitude is 5/64 of the original.
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