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 analyze the dimensions of its components:

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

Now, we compute the dimensions of \( \epsilon_0 E^2 \):

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

Multiplying these dimensions yields:

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

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

The determined dimensions are \([M L^{-1} T^{-2}]\).