Question
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.
Answers
drwls
Use Coulomb's law and perform a vector addition of forces.
JoshuaVitug
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
= -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
Anonymous
this is worst
i had type another ques and the given solu... is another 😤😤
i had type another ques and the given solu... is another 😤😤
Anonymous
So this is correct? Or what's the matter?
VHCP
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
= 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