Question
Use Symmetry or Die! Consider the fixed arrangement of changes on the square shown at right. All charges on a side are equally spaced and have a magnitude Q. We want to find the electric filed vector at the center of the square. A. Draw a Free-Body Diagram for a small positive test charge q placed at the center of the square that shows each of the 10 forces (i.e you should have 10 individual arrows). Make sure the relative lengths of the forces are correct and label the relevant angles with numeric values. Using a "brute force" method you would use this diagram to find the vector sum all 10 individual force component by component on the central test charge without any cleverness like making simplification due to symmetry. DO NOT DO THIS! Instead, if we are clever and consider symmetry before cranking through a problem, we can often simplify many things and save ourselves lots of work. b. In a set of concise statements explain all the simplifications that symmetry allows you to make in determining the net force on the test charge q. In as short a mathematical sum as possible (i.e 3 terms or less) write the terms that need to summed then show how they simplify. Give a final answer for the net force (magnitude and direction) on q in units of kQq/d^2. That is, your answer should be a number times kQq/d^2. Give its direction as well. d. What is the electric field at the center of the square? e. What is the electric field (not electric force) that an electron would feel if placed at the center of the square?
Explanation / Answer
b. As positive and -ve charges are on the diagonally opposite sides of the centre, both of them have same force in same direction
The x components of field will cancel out for the left 2 and right 2 positive charges (or the negative ones)
c. Field in y direction due to corner +ve charge = - kQcos(45)/d^2 = - 0.707kQ/d^2
theta = arctan(d2/d4) = 26.56 deg
Field in y direction due to positive charge below corner charge = -kQcos(theta)/((d/4)^2 + (d/2)^2) = -2.862kQ/d^2
Field in y direction due to centre positive charge = -4kQ/d^2
Net Field = - 4*0.707kQ/d^2 -4*2.862kQ/d^2 -2*4kQ/d^2 = -22.276kQ/d^2
Force = 22.276kQq/d^2 towards negative y axis
d. Electric field = 22.276kQ/d^2 towards -ve y axis
e. -22.276kQ/d^2
