Questions on Electrostatics:
1) We do not feel the gravitational forces between ourselves and the objects around us because these forces are extremely small. Electrical forces, in comparison, are extremely huge. Since we and the objects around us are composed of charged particles, why don’t we usually feel electrical forces?
Ans. Because to a large extent we are electrically neutral (the positive and negative charges are almost in perfect balance and cancel to give a net charge of zero).
2) When you remove your wool suit from the dry cleaner’s garment bag, the bag becomes positively charged. Explain how this occurs.
Ans. The physical act of separating the bag from the suit mechanically strips electrons off the bag. The electrons remain on the suit leaving it negatively charged whereas some of the atoms that are part of the bag are now missing their electrons and hence the bag has a net postive charge. The magnitude of charge on the bag and suit are equal but opposite as must be the case since charge is conserved.
3) How does a coulomb of charge compare to the charge of a single electron?
Ans. It’s enormous. It would take 1.6 x 1019 electrons to make up a coulomb of charge.
4) Two pellets, each with a charge of 1 microcoulomb (10-6 C), are located 3 cm (0.03 m) apart. What is the electric force between them? What mass object would experience this same force in the earth’s gravitational field?
Ans. Force = (9x109) x q1 x q2/d2. Substituting into this inverse-square law we get Force = (9x109) (10-6)( 10-6)/(.03)2 = 10 Newtons. An equivalent force from the earth gravitational field (say at the surface of the earth where acceleration is 10 m/s2) would involve a mass = Force/acceleration = 10 Newtons/(10 m/s2) = 1 kg.
5) A positively charged particle at a particular distance from a charged particle is attracted with a given electric force. By how much will the force change when the positively charged particle is three times as distant from the other particle? Five times as distant? One-quarter times as distant? (What law guides your answers?)
Ans. The electrical force for particles is governed by an inverse-square law. Hence if the distance is increased by a factor of 3 the force is reduced by a factor of 9. Increase the distance by a factor of 5, force is reduced by a factor of 25. Decrease the force by a factor of 4 and the force increases by a factor of 16 (1/(1/4)2) = 16)
6) What is the sign of charge of the other particle in the question above?
Ans. Oppositely charged particles attract so the other particle must be negatively charged.
7) Two equal charges exert equal forces on each other. What if one charge has twice the magnitude of the other. How do the forces they exert on each other compare?
Ans. Now here is a tricky question – the answer is that the forces they exert on each other are still equal because of Newton’s third law (for every action there is an equal and opposite reaction). The magnitude of the force in the latter case will be twice as large as in the former case, but the two charges each experience the same magnitude of force for each particular case. The same thing happens with the gravitational force between a golf ball and the earth. They exert equal but oppositely directed forces on each other. The golf ball is much less massive so we perceive it to be the only object of the pair to undergo acceleration.
Questions on Electric Current:
1) If an electric current flows from one object to another, what can we say about the relative magnitudes of the electric potentials of the two objects?
Ans. One object must be at a higher electric potential than the other. Like heat which only flows if there is a difference in temperature between two objects, current only flows if there is a difference in electrical potential.
2) Why are thick wires rather than thin wires usually used to carry large currents?
Ans. Thick wires have less resistance and so they will heat up less when a large current is passed through them. The resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area. If too much current is passed through a thin wire it will get hot enough to vaporize.
3) A 1-mile long copper wire has a resistance of 10 ohms. What will be its new resistance when it is shortened by (a) cutting it in half; (b) doubling it over and using it as “one” wire?
Ans. (a) 5 ohms (b) double it over and you provide two paths for the current to flow. Two equivalent paths result in half the original resistance. So the answer is again 5 ohms. Another way to think about this is to realize that doubling a wire over is equivalent to replacing the two parallel wires with one wire having twice the cross-sectional area. Since the resistance is inversely proportional to the cross-sectional area this results in half the resistance.
4) The damaging effects of electric shock result from the amount of current that flows in the body. Why, then, do we see signs that read “Danger – High Voltage” rather than “Danger-High Current”.
Ans. By Ohm’s Law, current = voltage/resistance. The current depends inversely on the resistance. In general, the resistance is unspecified and depends on the object and other conditions. Hence the resulting current cannot be defined nor assessed as to whether it is life threatening. For example, you can safely handle a high voltage source if you wear rubber gloves which are highly resistive. The current passing through you will be negligible. However, touching a high voltage source with wet hands (low resistance) can be lethal. The signs can only warn you of the potential of electric shock, the resistance of your body determines the current passing through it.
5) Why is the wingspan of birds a consideration in determining the spacing between parallel wires in a power line?
Ans. If the wingspan of birds is sufficient to span the wires that differ in electric potential, then the bird acts as a path for current to flow from the high voltage wire to the low voltage wire. Since the electric potential differences are huge in power lines, it is very likely that the bird will be cooked when it completes the circuit and the ground below will become littered with carcasses.
6) What changes occur in the line current when more devices are introduced in (a) a series circuit? (b) In a parallel circuit? Why are your answers different?
Ans. (a) line current drops because of the higher resistance (current = voltage/resistance).
(b) line current increases because there are additional paths for current to flow (lower resistance)
7) Why are devices in household circuits almost never connected in series?
Ans. If one of the devices were to fail in a catastrophic way (e.g. blow out like a fuse) then the circuit would be broken and current would not be able to flow to the other appliances and power them. By keeping devices in parallel, each device experience the same electric potential difference and can operate independently of the others.
8) Suppose you are refinishing the basement of your home and you decide to add more wall outlets. You tap into the power supply line. To prevent overloading in the circuits you will add a fuse or circuit breaker. Where and how (series or parallel) will you install the fuse?
Ans. The proper place to add the fuse is in series on the high voltage line before branching out the wires to the various outlets.
9) The wattage marked on a lightbulb is not an inherent property of the bulb but depends on the voltage to which it is connected, usually 110 or 120V. How many amperes flow through a 60-W bulb connected in a 120 V circuit?
Ans. Power= current x voltage. Rearranging terms we find that current = power/voltage = 60W/120V = 0.5 Amps.
10) Rearrange the Current = voltage/resistance to express resistance in terms of current and voltage. Then solve the following: An air conditioner in a 120 V circuit has a current rating of 20 A. What is the resistance of the air conditioner (how many ohms)?
Ans. Resistance = Voltage/current = 120V/20A = 6 ohms.
11) A 4 Watt night light is plugged into a 120 V circuit and operates continuously for 1 year. Find the following: (a) the current it draws, (b) the resistance of the filament, (c) the energy consumed in a year, and (d) the cost of its operation for a year at the utility rate of 10 cents per kilowatt-hour.
Ans. (a) current = power/voltage = 4W/120V = 1/30 th of an amp or 0.033 Amps
(b) resistance = voltage/current = 120V/(1/30) = 3600 ohms
(c) energy = power x time = 4 Watts x time = 4 Joules/sec x 60s/minute x 60minutes/hour x 24hours/day x 365days/year which is approximately equal to 126,000,000 Watts.
(d) 1 kilowatt-hour = 1000 Watts x 60sec/minute x 60 minutes/hour = 3600000 Joules
number of kilowatt-hours used is equal to the total number of joules used divided by the number of Joules in a kilowatt-hour. This is equal to 126,000,000/3,600,000 = 35 kilowatt-hours. At 10 cents per kilowatt-hour the total annual cost for operating the night light is $3.50.