Question

Suppose Q1 (0.0, - 3.0 M, 0.0) = 4.0 nC, Q2 (0.0, 3.0 m, 0.0) = 4.0 nC, and Q3 (4.0 m, 0.0, 0.0) = 1.0 nC.

a) Find the total force acting on the charge Q3.
b) Repeat the problem after changing the charge of Q2 to -4.0 nC.
c) Find the electric field intensity at the location of Q3 for parts a) and b)

Answer 1

Answer:

Explanation:

Force on Q₃ due to charge Q₁

= 9 x 10⁹x 4 x 10⁻⁹ x1 x 10⁻⁹ /  5²

= 1.44 x 10⁻⁹ N

Force due to Q₂ will also be 1.44 x 10⁻⁹ N

component of these forces along x axis

-= 2 x 1.44 x 10⁻⁹ cosθ

= 2.88 x  10⁻⁹  x 4/5

= 2.30x10⁻⁹  N along x axis.

The y-component will calcel out.

b ) In this case , Q₁ will repel and Q₂ will attract.

In this case Q₁ will repel and Q₂ will attract. Component along y - axis will be same as earlier one or 2.30x10⁻⁹  N . Component along x axis will cancel out.

c ) Electric field in case 1 and case 2 will be

= 2.30x10⁻⁹  / 1 x 10⁻⁹

= 2.3 N / C , because field is force per unit charge. The sane field will be in case 2 .