#$&* course Phy 242 027.
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Given Solution: STUDENT RESPONSE FOLLOWED BY SOLUTION: The problem said that V was 0 at d = inifinity, which I understnad to mean that as we approach the raindrop from infinity, the potential differencegrows from 0, to some amount at the surface of the raindrop. Because water molecules are more positive on one side that the other, they tend to align in a certain direction. Since positive charges tend to drift toward negative charge, I would think that the raindrop, with its overall negative charge, has molecules arranged so that their more positive sides are pointing toward the center and negative sides will be alighed along the surface of the raindrop. Probably all wrong. I tried several differnet integrand configuraitons but never found one that gave me an answer in volts. SOLUTION: You will have charge Q = -1.2 * 10^-12 C on the surface of a sphere of radius .00065 m. The field is therefore E = k Q / r^2 = 9 * 10^9 N m^2 / C^2 * (-1.2 * 10^-12 C) / r^2 = -1.08 * 10^-2 N m^2 / C / r^2. Integrating the field from infinity to .00065 m we get (-1.08 * 10^-2 N m^2 / C) / (.00065 m) = -16.6 N m / C = -16.6 V. If two such drops merge they form a sphere with twice the volume and hence 2^(1/3) times the radius, and twice the charge. The surface potential is proportional to charge and inversely proportional to volume. So the surface potential will be 2 / 2^(1/3) = 2^(2/3) times as great as before. The surface potential is therefore 16.6 V * 2^(2/3) = -26.4 volts, approx.. ** STUDENT QUESTION: I knew that my answer was off by some factor because the E decreased from when it was just one raindrop. I didn’t get that you would multiply it by two because the volume increased. ??? Can you explain why you would use the ratio of volume to radius increase in order to get the new E??? INSTRUCTOR RESPONSE: E depends on the total charge and the radius. When the two drops merge, their charges combine. This gives you double the charge compared to a single drop. We don't care about the volume, we care about the radius. However we know what happens to the volume: it doubles. So we use what we know about the volume to determine what happens to the radius: A sphere with twice the volume of another has 2^(1/3) times the radius. We end up with double the charge on a sphere with 2^(1/3) times the radius. Since the potential is proportional to the charge and inversely proportional to the radius, the potential changes by factor 2^(2/3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):goooooddd ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: