Answer to Question #91758 in Electric Circuits for Shali

Question

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.

Answer 1

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.