# A charge of -3.00 ?C is fixed at the center of ac...

A charge of -3.00 ?C is fixed at the center of acompass. Two additional charges are fixed on the circle ofthe compass (r= 0.100 m). The charges on the circle are -4.00?C at the position due north and +5.00 ?C at the positiondue east. What is the magnitude and direction of the netelectrostatic force acting on the charge at the center? Specifiy the direction relative to due east,

## Answers ( 1 )

1. For this problem, you have to plug in both charges (i.e theones surrounding the center charge, w/ thecenter charge) with Coulomb's Law:
F1 =k(3E-6C)(4E-6C)/(0.125m)^2
=?
F2 = k(3E-6C)(5E-6C)/(0.125m)^2
=?
With the 2 values you get here, use the Pythagorean Theorem(a^2+b^2 = c^2) to find the c value.
That c value (F3) is your electrostatic force.
To find the direction due east, use one of your trig. functionto get the angle, so for example:
inverse sin (F1/F3) = the angle
Don't forget you're looking for the angle due east...so it maybe negative...
Good luck!