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

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 \([ML^{-1}T^{-2}]\).