Question

The unit of permittivity of vacuum is :

• A. Newton m\(^2\)/coulomb\(^2\)
• B. coulomb\(^2\)/Newton m\(^2\)
• C. Newton/coulomb
• D. Newton volt/m\(^2\)

Hint:

A common mistake is to confuse the units of \(\epsilon_0\) with the units of the electrostatic constant \(k = 1/(4\pi\epsilon_0)\). The units of k are N m\(^2\)/C\(^2\), which is the reciprocal of the units of \(\epsilon_0\). Option (A) is the unit for k.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The permittivity of vacuum, denoted by \(\epsilon_0\), is a physical constant that represents the capability of a vacuum to permit electric field lines. Its unit can be derived from any equation where it appears, most commonly Coulomb's Law.
Step 2: Key Formula or Approach:
Coulomb's Law for the electrostatic force (F) between two point charges (q\(_1\) and q\(_2\)) separated by a distance (r) is: \[ F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \] We can rearrange this formula to solve for \(\epsilon_0\) and then find its units.
Step 3: Detailed Explanation:
Rearranging Coulomb's Law to solve for \(\epsilon_0\): \[ \epsilon_0 = \frac{1}{4\pi F} \frac{q_1 q_2}{r^2} \] Now, let's substitute the SI units for each quantity (the constant \(4\pi\) is dimensionless):
Force (F): Newton (N)
Charge (q\(_1\), q\(_2\)): Coulomb (C)
Distance (r): meter (m)
Substituting these into the rearranged formula: \[ \text{Unit of } \epsilon_0 = \frac{1}{\text{N}} \frac{\text{C} . \text{C}}{\text{m}^2} = \frac{\text{C}^2}{\text{N} . \text{m}^2} \] This can be written as coulomb\(^2\)/Newton m\(^2\).
Step 4: Final Answer:
The unit of permittivity of vacuum is coulomb\(^2\)/Newton m\(^2\).