QuestionJune 1, 2025

What would be the magnitude of the resulting force between an object of charge 75mu C and another object of charge 50mu C that are placed 4 m apart? 3.75 N 3.14 N 1.40 N 1.69 N

What would be the magnitude of the resulting force between an object of charge 75mu C and another object of charge 50mu C that are placed 4 m apart? 3.75 N 3.14 N 1.40 N 1.69 N
What would be the magnitude of the resulting force between an object of charge 75mu C
and
another object of charge 50mu C that are placed 4 m apart?
3.75 N
3.14 N
1.40 N
1.69 N

Solution
4.5(318 votes)

Answer

2.11 N Explanation 1. Identify the formula Use Coulomb's Law: **F = k \frac{|q_1 q_2|}{r^2}**, where k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2, q_1 = 75 \times 10^{-6} \, \text{C}, q_2 = 50 \times 10^{-6} \, \text{C}, and r = 4 \, \text{m}. 2. Substitute values into the formula F = 8.99 \times 10^9 \times \frac{(75 \times 10^{-6})(50 \times 10^{-6})}{4^2} 3. Calculate the force F = 8.99 \times 10^9 \times \frac{3750 \times 10^{-12}}{16} 4. Simplify the expression F = 8.99 \times 10^9 \times 234.375 \times 10^{-12} 5. Final calculation F = 2.10703125 \, \text{N}

Explanation

1. Identify the formula<br /> Use Coulomb's Law: **$F = k \frac{|q_1 q_2|}{r^2}$**, where $k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$, $q_1 = 75 \times 10^{-6} \, \text{C}$, $q_2 = 50 \times 10^{-6} \, \text{C}$, and $r = 4 \, \text{m}$.<br /><br />2. Substitute values into the formula<br /> $F = 8.99 \times 10^9 \times \frac{(75 \times 10^{-6})(50 \times 10^{-6})}{4^2}$<br /><br />3. Calculate the force<br /> $F = 8.99 \times 10^9 \times \frac{3750 \times 10^{-12}}{16}$<br /><br />4. Simplify the expression<br /> $F = 8.99 \times 10^9 \times 234.375 \times 10^{-12}$<br /><br />5. Final calculation<br /> $F = 2.10703125 \, \text{N}$
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