Rankers Physics

Coulomb's Law: Practice Problem & Solution

6. The unit of permittivity of free space $\epsilon_0$ is: (2004)
$\text{Newton metre}^2/\text{Coulomb}^2$
$\text{Coulomb}^2/\text{Newton metre}^2$
$\text{Coulomb}^2/(\text{Newton metre})^2$
$\text{Coulomb}/\text{Newton metre}$

Solution Explained:

To solve this problem, we apply the core principles of Coulomb's Law. Understanding the underlying formula is key to arriving at the correct answer below:

From Coulomb's Law, $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$. Rearranging gives $\epsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$. The unit is $\text{Coulomb}^2/\text{Newton metre}^2$.

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