Four objects, each with charge +q, are held fixed on a square with sides of length d, as shown in Figure 18.66. Objects X and Z are at the midpoints of the sides of the square. The electrostatic force exerted by object W on object X is F. What is the magnitude of force exerted by object W on Z? (a) F/7
(b) F/5
(c) F/3
(d) F/2

Video Transcript

This is College Physics Answers with Shaun Dychko. We're given a square with side length d, and four charges on it of equal magnitude, q. Charges x and z are at the mid-point of their respective sides. So the force between w and x we're told is F. We'll call the distance between them r x. The question is what is the force on charge z? The separation between w and z is r subscript z. So the force on z is going to be k q squared over r z squared and the force on x which is just called F is going to be k q squared over r x squared. Now the question is how do we relate r x and r z? If we do that we'll be able to relate these two forces. So r x is d over two because it's at the mid-point of the side and r z is going to be the hypotenuse of this triangle here, and this triangle has a side length of d on one side and d over two on the other side. So r z is going to be the square root of the sum of the squares of the legs of this right triangle. So that's square root of d squared plus d over two squared. That's d squared plus d squared over four. This is four over four d squared you could say, and that makes a total of five over four d squared under the square root sign. Then we'll do the square rooting. So we get square root five over four times d which is root five over two, times d. d over two is r x so we can substitute r x in place of d over two. So r z is root five times r x. Now that means we can replace r z with this. So we've done that here. So F z is k q squared over r z which we've replaced with root five times r x squared and that makes one fifth k q squared over r x squared and k q squared over r x squared is F. So F z is one fifth F. So the answer is B.