Two charges are fixed on the x axis: one with a charge q1=5.0x10^-6 C at x1 =1.0

Question

Two charges are fixed on the x axis: one with a charge q1=5.0x10^-6 C at x1 =1.00 m, and the other with a charge of q2=3.0x10^-6 at x2=1.5 m. Find the magnitudes of the three charges, and the force exerted on charge q=5.0x10^-6 C that is placed at the origin. Two charges are fixed on the x axis: one with a charge q1=5.0x10^-6 C at x1 =1.00 m, and the other with a charge of q2=3.0x10^-6 at x2=1.5 m. Find the magnitudes of the three charges, and the force exerted on charge q=5.0x10^-6 C that is placed at the origin.

Explanation / Answer

i think correct question is

Two charges are fixed on the x axis: one with a charge of q1 = 5.00uC at x1 = -1.00m and the other with a charge of q2 = 3.00uC at x2 = 1.50m . Find the force F exerted on a charge q = -5.00uC placed at the origin (x=0).What are the magnitudes of the the three charges q, q1, and q2?

Answer:

q1 = 5x10^-6 C

q2 = 3x10^-6 C

q = -5 x10^-6 C

x1 = 1 m ( take the magnitude of the distance)

x2 = 1.5 m

Using coulombs law ,

F1 = 9*10^9*q1*q/1 = 225*10^-3 N ( to the left as it is the force of attraction )

F2 = 9*10^9*q2*q/1.5^2) = 9*10^-3 ( 5*3 /1.5^2) = 60*10^-3 N ( to the right as it is the force of attraction )

Net force on q by q1 and q2 is,

F = F2 - F1 = 60*10^-3 - 225*10^-3 = -165*10^-3 N = - 0.165 N ,

The net force is towards the left.

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