Answer to Question #91758 in Electric Circuits for Shali

Question #91758
1. A charge q = - 1.0 C is placed at (0, 0) m while another charge Q = 3.0 C is placed at (0.15, 0) m. Calculate the magnitude of the electric force acting on q.
1
Expert's answer
2019-07-17T10:12:22-0400

The magnitude of the electrostatic force F between point charges q and Q separated by a distance r is given by Coulomb’s law


"F=k\\frac{\\left| q\\cdot Q \\right|}{{{r}^{2}}}\\,\\,\\,\\,\\,\\,\\left( 1 \\right)"


where constant k is equal to 9·109 N·m·C-2.

In accordance with the Newton’s third law (every force exerted creates an equal and opposite force) the force on q is equal in magnitude and opposite in direction to the force it exerts on Q, so the magnitude of the electric force acting on q is (1).

Find the distance between charges. Both charges lie on the x axis, so the distance between them is equal to the difference between their abscissas, r=0.15 m.

Substitute the known values in (1)


"F=9\\cdot {{10}^{9}}\\frac{N\\cdot {{m}^{2}}}{{{C}^{2}}}\\cdot \\frac{\\left| -1.0\\cdot {{10}^{-6}}C\\cdot 3.0\\cdot {{10}^{-6}}C \\right|}{{{\\left( 0.15\\,m \\right)}^{2}}}=1.2\\,N"


The charges are opposite in sign, so this is an attractive force.


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