High School

Kalyan Ramji Sain, of India, had a mustache that measured 3.39 m from end to end in 1993. Suppose two charges, [tex]q[/tex] and [tex]3q[/tex], are placed 3.39 m apart. If the magnitude of the electric force between the charges is [tex]2.4 \times 10^{-6} \, \text{N}[/tex], what is the value of [tex]q[/tex]?

Answer :

Final answer:

The value of q is approximately 2.01 × 10^(-19) C.

Explanation:

The magnitude of the electric force between two charges can be calculated using the formula:

F = k * |q1| * |q2| / r^2

In this case, we are given the magnitude of the force F = 2.4 × 10^(-6) N, the distance between the charges r = 3.39 m, and the ratio of the charges q / 3q = 1 / 3. Using these values, we can solve for q.

Substituting the given values into the formula and solving for q, we get:

2.4 × 10^(-6) N = (8.99 × 10^9 N m^2/C^2) * |q| * |3q| / (3.39 m)^2

Simplifying the equation, we find:

q = 2.4 × 10^(-6) N * (3.39 m)^2 / ((8.99 × 10^9 N m^2/C^2) * 3 * 9)

Calculating the value of q, we get q ≈ 2.01 × 10^(-19) C.

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