Change the chapter
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?
  1. F/7
  2. F/5
  3. F/3
  4. F/2
Question Image
<b>Figure 18.66</b> Equal charges placed on a square.
Figure 18.66 Equal charges placed on a square.
Question by OpenStax is licensed under CC BY 4.0.
Final Answer


Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 18, Problem 39 (Test Prep for AP® Courses) (2:12)

Sign up to view this solution video!

View sample solution
Video Transcript

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