Static Electricity Legacy Problem #12 Guided Solution

There are a collection of habits that an effective problem solver typically employs in order to approach a problem in physics. The more difficult the problem, the more important that the problem solver uses these effective habits. One of the most important habits is a careful reading of the problem and a visualization of the situation. Here in this problem, we read about two bowling balls which are in a vertical cylinder. I actually tend to begin to draw a diagram in which I draw or sketch a little vertical cylinder and I put two object circles, bowling balls, within that cylinder. What I know about these bowling balls is that they are charged with such a charge that the top ball begins to levitate above the bottom ball. This occurs because the force of electrical repulsion of the top ball by the bottom ball is equal to the force of the gravity of being attracted to the earth. As such, I can begin this problem by stating the equation that the F electrical on the top bowling ball is equal to the F gravity on the top bowling ball. Doing so allows me to write this statement that the k times the q of the top ball times the q of the bottom ball divided by the distance squared is equal to the mass of that top ball multiplied by g, 9.8 newtons per kilogram. In this problem, I wish to calculate the q value, the quantity of charge on both the top and the bottom bowling ball. I am told that they are charged identically, that one ball has the same charge as the other. Thus, the q of the top ball times the q of the bottom ball is simply equal to q times q or q squared. So my equation now becomes k times q squared over the distance squared that separates them is equal to mass of the top ball multiplied by 9.8 newtons per kilogram. Now I should be able to solve for the quantity of charge on the bowling ball because I know the distance of separation and I know the mass of the bowling ball. The mass of the bowling ball is 7.25 kilograms and the distance of separation can be calculated if I use the information given about the distance that separates the ball's surfaces as well as the diameter of the bowling ball. Now here is where that initial diagram that I drew comes in so handy because I drew my two bowling balls in this vertical cylinder such that they were separated by a distance of 25 centimeters. That's between their nearest surfaces. The D and Coulomb's law equation is actually the distance between the two centers of the ball. Now each ball has a diameter of 21.8 centimeters. If I have that I can get my radius and that would be a radius value of 10.9 centimeters. So the distance between the center of one ball and the center of the other ball can be worked out to be 10.9 centimeters plus 25 centimeters plus 10.9 centimeters. Adding this up gives me a value of 46.8 centimeters. That's the distance from center to center. One further complication of this problem is that's a distance in centimeters and as I'm using the Coulomb's law equation I need to substitute values of D in units of meters. That's because my value of K is 8.99 Newtons times a meter squared per Coulomb squared. So in order to get my units to cancel appropriately I need to make sure my D value is in units of meters. So I'm going to take the decimal place on 46.8 and I'm going to move it two places to the left. Doing so gives me 0.468 meters and I can substitute that into my Coulomb's law equation and proceed to solve for the value of Q. Now when I do that I get about 4.16 times 10 to the negative 5th Coulomb's which is a huge amount of charge. If I wish I could convert that to microcoulomb by simply multiplying by 10 to the 6th since there's 10 to the 6th of these little microcoulombs in every one Coulomb.