three point charges are placed at the following points on the x-axis: +2µC at x=0, -3µC at x=40 cm, -5µC at x=120 cm. Find the force (a) on the -3µC charge (b)on the -5µC charge.

Answer 1

Use Coulomb's law and perform a vector addition of forces.

Answer 2

Fe= (-9.0 * 10^9) * (2.0 * 10^-6)(3.0 * 10^-6) / 0.40^2
= -54 * 10^-3 / 1.6 * 10^-1
= -0.3375 N

Fe= (-9.0 * 10^9) * (-5.0 * 10^-6)(3.0 * 10^-6) / 0.80^2
= 135 * 10^-3 / 6.4 * 10^-1
= -0.2109 N

a = -0.55N

Fe= (9.0 * 10^9) * (2.0 * 10^-6)(5.0 * 10^-6) / 1.20^2
= 90 * 10^-3 / 1.44 * 10^1
= 0.0625 N

Fe= (9.0 * 10^9) * (-5.0 * 10^-6)(3.0 * 10^-6) / 0.80^2
= 135 * 10^-3 / 6.4 * 10^-1
= 0.2109 N

=0.27N

Answer 3

this is worst

i had type another ques and the given solu... is another 😤😤

Answer 4

So this is correct? Or what's the matter?

Answer 5

Fe= (9.0 * 10^9) * (-5.0 * 10^-6)(3.0 * 10^-6) / 0.80^2
= 135 * 10^-3 / 6.4 * 10^-1
= 0.2109 N

Fe= (9.0 * 10^9) * (2.0 * 10^-6)(5.0 * 10^-6) / 1.20^2
= 90 * 10^-3 / 1.44 * 10^1
= 0.0625 N

B) 0.2109375 - 0.0625 = 0.1484375