Point charge 1.5?C is located at x = 0, y = 0.30 m , point charge...

Point charge 1.5?C is located at x = 0, y = 0.30 m , point charge -1.5?C is located at x = 0 y = -0.30 m . What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0?C at x = 0.40 m , y = 0?

Expert Solution

Draw the charges! Call the one at (0, 0.3) charge 1; call the one at (0, -0.3) charge 2, and the one out at (0.4, 0) charge 3.
Now, charge 1 exerts a force pushing charge 3 away from it, at an angle of arcsin(3/5) ~~ 36.87deg. below the x-axis. Charge 2 exerts a force attracting charge 3 to it, but it is also inclined at 36.87 deg. below the axis, and they're of the same magnitude (from q1*q2/r^2), so the horizontal components cancel out. Therefore, one only has to find twice the vertical component of one of the forces to find the net force.
This is given by Fvert = (8.9876e9 N*m^2/C^2)*(1.5e-6C * 5.0e-6C)/(0.5m)^2 * (3/5). The first of these factors is Coulomb's constant, in MKS units. The second of the factors is due to q1*q2 in Coulomb's Law; the third factor is (1/r^2) in Coulomb's Law; the fourth factor is due to the vector nature of the force, and is just the vertical component. Thus, Fvert = 0.162 N, and twice this (from each charge) is about 0.32 N.

genius_generous answered 2 years ago

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