4 20 points Two 2 mu C point charges are located on the xaxi

4 (20 points) Two 2 mu C point charges are located on the x-axis. One is at x =1.0 m and the Other is at x= -1.0 m. (a) Determine the electric field on the y axis at = 0.50 m. (b) Calculate the electric force on a -3 mu C charge placed on the y axis at y = 0.50 m.

Solution

Part A)

Start with E = kq/r²

The distance r, by the Pythagorean Theorem is...

r² = 1² + .5²

r = 1.12 m

E = (9 X 10⁹)(2 X 10^-6)/(1.12)²

E = 14349 V/m

The y compenets will cancel, but the x components will add

We need the angle using the tangent function

tan(angle) = .5/1

angle = 26.6^o

The E field in the x direction will be 2(14349)(cos 26.6)

E = 25669 V/m (Round to 2.57 X 10⁴ V/m)

Also NOTE: V/m is the same as N/C depending on the units your book wants you to use. They mean the same thing)

Part B)

F = qE

F = (3 X 10^-6)(25669)

F = .077 N (7.70 X 10^-2 N)