Cycle II AP Physics
John Dewey High School
Mr. Klimetz
Electric Forces and Electric Fields
Text Chapter 19
Solution to Problem 65
In the rectangle in the drawing, a charge is to be placed at the empty corner to make the net force on the charge at corner A point along the vertical direction. What charge (magnitude and algebraic sign) must be placed to the empty corner?
A
+3 uC
+3 uC
+3 uC
Q uC
Solution:
Given:

Width of Rectangle = d
Length of Rectangle = 4d
Corner Charges (q) = +3 uC
d
4d
Equation:

Fe = kq1q2/(dexp2)
where k = 9.0 x 10exp9 N-mexp2/Cexp2, d = distance, m, q1 and q2 are charge values, C.

Hypotenuse = 4.12d
4d
14 degree angle: Leg 4d to Hypotenuse
Solving for the angle between Leg 4d and the Hypotenuse
tanexp-1 (d/4d) = 14 degrees
In order for the force on A to be
restricted to the vertical (y-axis)
direction, we must counter the force of repulsion between charge A and the charge at the lower right-hand corner of the rectangle, oriented along Leg 4d. In other words, the net force felt by charge A along Leg 4d must be the equilibrant of the forces exerted along the hypotenuse (the line between charge A and the unknown charge) and the unchanged vertical component of force exerted along Leg d (the repulsive force between A and the charge at the upper left-hand corner of the rectangle).

Therefore,

= force along Leg 4d/force along hypotenuse

= FLeg4/Fhyp

[(k x 3 uC x 3 uC)/16dexp2]
[(k x 3 uC x Q)/17dexp2]

where Q is the unknown charge.
Therefore

cos14 degrees = (3 uC/16)/(Q/17)

or

Q = (0.19) x (17) /(0.97) uC

then

Q = -3.3 uC
=
d
Q