Question

select the correct answer.

two balloons having the same charge of $1.2 \times 10^{-6}$ coulombs each are kept $5.0 \times 10^{-1}$ meters apart. what is the magnitude of the repulsive force between them? ($k = 9.0 \times 10^9$ newton·meter²/coulombs²)

\\(\bigcirc\\) a. $1.2 \times 10^{-2}$ newtons

\\(\bigcirc\\) b. $3.2 \times 10^{-3}$ newtons

\\(\bigcirc\\) c. $6.5 \times 10^{-3}$ newtons

\\(\bigcirc\\) d. $3.0 \times 10^{-4}$ newtons

\\(\bigcirc\\) e. $5.2 \times 10^{-2}$ newtons

Explanation:

Step1: Recall Coulomb's Law

Coulomb's Law is given by \( F = k\frac{q_1q_2}{r^2} \), where \( k = 9.0\times 10^{9}\ N\cdot m^{2}/C^{2} \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges. Here, \( q_1 = q_2 = 1.2\times 10^{-6}\ C \) and \( r = 5.0\times 10^{-1}\ m \).

Step2: Substitute the values into the formula

Substitute \( k = 9.0\times 10^{9} \), \( q_1 = 1.2\times 10^{-6} \), \( q_2 = 1.2\times 10^{-6} \), and \( r = 5.0\times 10^{-1} \) into the formula:

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Answer:

E. \( 5.2\times 10^{-2} \) newtons