Question

If \( \epsilon_0 \) is the permittivity of free space and \( E \) is the electric field, then \( \epsilon_0 E^2 \) has the dimensions:

• A. \([M^0 L^{-2} T A]\)
• B. \([M L^{-1} T^{-2}]\)
• C. \([M^{-1} L^{-3} T^4 A^2]\)
• D. \([M L^2 T^{-2}]\)

Answer 1

The Correct Option is B

To determine the dimensions of \( \epsilon_0 E^2 \), we need to understand the dimensions of each component:

1. The permittivity of free space, denoted by \( \epsilon_0 \), has dimensions of \([M^{-1} L^{-3} T^4 A^2]\). This is derived from the relation for Coulomb's law, or by using the formula for capacitance \( C = \frac{\epsilon_0 A}{d} \), where A is area and d is the distance.
2. The electric field \( E \) has dimensions of \([M L T^{-3} A^{-1}]\). This arises from the definition \( E = F/q \), where F is the force with dimensions \([M L T^{-2}]\) and \( q \) is the charge with dimensions \([A T]\).

Now, let's calculate the dimensions of \( \epsilon_0 E^2 \):

• \( E^2 \) has the dimensions of \([M L T^{-3} A^{-1}]^2 = [M^2 L^2 T^{-6} A^{-2}]\).
• Thus, \( \epsilon_0 E^2 \) has dimensions \([M^{-1} L^{-3} T^4 A^2] \times [M^2 L^2 T^{-6} A^{-2}]\).

Performing dimensional multiplication, we have:

• Mass: \((-1) + 2 = 1\)
• Length: \((-3) + 2 = -1\)
• Time: \(4 - 6 = -2\)
• Current: \(2 - 2 = 0\)

Thus, the dimensions of \( \epsilon_0 E^2 \) are \([M L^{-1} T^{-2}]\).

The correct answer is therefore: \([M L^{-1} T^{-2}]\).

Answer 2

The electric field is given by:
\[E = \frac{KQ}{R^2}.\]
Substituting \(K = \frac{1}{4\pi \epsilon_0}\), we get:
\[E = \frac{Q}{4\pi \epsilon_0 R^2}.\]
From this, the permittivity of free space (\(\epsilon_0\)) can be expressed as:
\[\epsilon_0 = \frac{Q}{4\pi R^2 E}.\]
Now, calculate \(\epsilon_0 E^2\):
\[\epsilon_0 E^2 = \frac{Q}{4\pi R^2 E} \cdot E^2 = \frac{QE}{4\pi R^2}.\]
Analyzing the dimensional formula:
\[[\epsilon_0 E^2] = \frac{[Q][E]}{[R^2]}.\]
Substituting the dimensional formulas:
\[[Q] = [W], \quad [E] = \frac{[W]}{[R^2][Q]}.\]
\[[\epsilon_0 E^2] = \frac{[W]}{[R^3]} = \frac{ML^2T^{-2}}{L^3}.\]
Simplifying:
\[[\epsilon_0 E^2] = [ML^{-1}T^{-2}].\]
Thus, the dimensions of \(\epsilon_0 E^2\) are \([ML^{-1}T^{-2}]\).