Question

Two charged particles are placed 2.0 meters apart. The first charge is +2.0 E-6 C, and the second charge is +4.0 E-6 C. What is the electrical force between the two charges? (5 points)
(k = 9.0 E9 Nm2/c2)
+1.8 E-2 N and it is repulsive
+3.6 E-2 N and it is repulsive
+4.5 E-2 N and it is attractive
+9.0 E-2 N and it is attractive

Answer 1

The electrical force between the two charges can be calculated as +1.8×10⁻² N and its repulsive.

To find the electrical force, distance = 2 meters

charge q1 = 2×10⁻⁶ C

charge q2 = 4×10⁻⁶ C

Define coulomb's law and solve with formula.

         The force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between them.

Formula can be written as,

          F = K ( q1q2 / r² )

F - electric force

k - Coulomb constant

q1, q2 - charges

r - distance of separation

Substituting the values in the formula,

         F = 9 × 10⁹ Nm²/C² ( (2×10⁻⁶ C ×  4×10⁻⁶ C) / (2²))

            = 0.0179751

         F = 1.8 × 10 ⁻² N.

As both the charges q1 and q2 are positive, the charges gets repulsive.

SO, the correct answer is Option A.

Learn more about coulomb's law,

brainly.com/question/506926

#SPJ1