Question

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°).

Magnitude
_________N
Direction
_________°



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Physics - Bobpursley, Thursday, August 30, 2007 at 1:52am
assume the charges don't move. Work this as a vector problem. Find the force S due to the center charge and the N charge, and the Force E due to the center charge and the E charge. Add those as vectors

Answer 1

Use Coulombs' Law:
F=k(q1)(q2)/(r)^2
***(q1 and q2 are the absolute values--so the sign is positive even if the given value is negative)
***K is the constant: 8.99 x 10^9N*m^2/C^2
***r is the radius

The question wants you to find the net force acting on the center particle--so Fnet= Force on center by the particle due east + the force on the center by the particle due north

Use Coulomb's Law to find the individual forces that the particles exert: (change uC to C for the equation)

Force on center by East= 8.99x10^9 (3.0x10^-6)(5.0x10^-6)/(0.135)^2
***There is no Y component for this force
*** The center charge is - and the east charge is + so the particles attract and the east charge has a force in the negative direction

Force on center by North: 8.99x10^9 (3.0x10^-6)(3.20x10^-6)/ (0.135)^2
***There is no X component for this force
***Like charges repel, so the force will be away from the center in the positive y direction.

Now, use A^2+B^2=C^2 to solve for C (this will be the vector sum of the forces) A=Force on center by North and B=Force on center by East (The answers you obtain from the two previous equations)

C will be your answer. (I don't have a calculator)

Direction: tan^-1 (opposite or y vector/adjacent or x vector)