Question

Two 45.0 x 10^-9 C point charges are located on the x axis. One is at x = 0.25 m, and the other at x = - 0.25 m.

a) A third identical charge is placed on the y axis at y = 0.25 m. Find the magnitude of the force acting on this third charge? Answer in Newtons.

b) Now the third identical charge is placed on the y axis at y = 2 x 0.25 m. Find the magnitude of the force acting on this third charge? Answer in Newtons.

Answer 1

Part A)

The formula that applies here is F = kqq/r²

We need the distance from the x locations to the y location from the Pythagorean Theorem

r² = (.25)² + (.25)²

r² = .125

So F = (9 X 10⁹)(45 X 10^-9)²/(.125)

F = 1.458 X 10^-4 N

This force has x and y components. They will be both the same since the angle is 45^o

F_{x and y} = (1.458 X 10^-4)(cos 45) = 1.03 X 10^-4 N

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

The net force is 2(1.03 X 10^-4) = 2.06 X 10^-4 N upward

Part B)

We need the new distance from the x locations to the new y location from the Pythagorean Theorem

r² = (.5)² + (.25)²

r² = .3125

So F = (9 X 10⁹)(45 X 10^-9)²/(.3125)

F = 5.832 X 10^-5 N

This force has x and y components. The x component will cancel, and the y components will add

We only need the y component

The angle formed is from teh tangent function

tan(angle) = .5/.25

angle = 63.43^o

The y component is (5.832 X 10^-5)(sin 63.43) = 5.216 X 10^-5 N

The net force is 2(5.216 X 10^-5) = 1.04 X 10^-4 N upward