PHYS208 Fundamentals of Physics II

Quiz 1 -- Solution

2 charges on a clock face

Two charges are fixed on the perimeter of a circular clock face.
+Q is placed at 2 o'clock; +2Q is at 12 o'clock.

Where should a third charge be placed on the perimeter so that the electric field is zero at the center (point P)?

What relative charge strength is required at that point for the electric field to be zero?


Working with dimensionless units is a straightforward way to proceed since all three charges are the same distance from the point of evaluation. Assign one unit of electric field to correspond to one unit of charge situated on the perimeter of the clock face. Considering a positive test charge at point P, we find one unit of electric field pointing toward 8 o'clock contributed from +Q and two units of electric field pointing toward 6 o'clock from +2Q. In the drawing below, I show two units of electric field spanning the distance from the center to the perimeter.

2 contributions to the electric field

The two contributions must be added vectorially to find the resultant electric field. It appears to put somewhat past 6:30 on the clock.

Resultant electric field

Resolving into components, observing that each hour on a clockface is separated by 30 degrees:

  Horizontal:  Vertical: 
+Q 0.866 left 0.500 down
+2Q --- 2.000 down

resultant:  0.866 left 2.500 down

Combining to find the magnitude of the resultant vector yields 2.646 units of electric field at a direction of 19.1 degrees clockwise from 6 o'clock (since the tangent of theta is 0.866/2.500 = 0.346.) This angle corresponds to 38 minutes, 13 seconds past 6 o'clock.

A third charge is to be added on the perimeter to balance the resultant of the two charges above. Since each unit of electric field corresponds to a unit of charge, a charge of 2.646Q will be required.

There are two possibilities for counteracting the other two charges:

+2.646Q at 6:38  OR  -2.646Q at 12:38.

Last updated Feb. 21, 1998.
Copyright George Watson, Univ. of Delaware, 1998.