A charge of -3.00 µC is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass (radius = 0.135 m). The charges on the circle are -3.20 µC at the position due north and +5.00 µC at the position due east. What is the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east (0°).
Physics - drwls, Saturday, September 1, 2007 at 10:24am
There are two forces acting on the charge at the center.
Due to the -3.2 uC charge at the north point, there is a repulsion force of
F1 = k*(3.0)*(3.2)*10^-12/(0.135)^2
and it points to the south.
Due to the +5.0 uC charge at the east point, there is an attraction force of
F2 = k*(5.0)(3.2)*10^-12/(0.135)^2
and it points to the east.
k is the Coulomb's Law constant,
9.0 x 10^9 N • m2 / C
Do the multiplication and add the two forces vectorially. The direction of the resultant force will be
arctan 5.0/3.0 = 59.0 degrees south of (clockwise from) east, since 5/3 is the ratio of the two component forces.