Question

The dimension of $\frac{1}{2}\varepsilon _0 E^2$, where $\varepsilon _0$ is permittivity of free space and $E$ is electric field, is

• A. $ML^2T^{-2}$
• B. $ML^{-1}T^{-2}$
• C. $ML^2T^{-1}$
• D. $MLT^{-1}$

Answer 1

The Correct Option is B

To determine the dimension of the expression $\frac{1}{2}\varepsilon _0 E^2$, we need to understand the dimensions of each component in this expression.

The expression $\frac{1}{2}\varepsilon _0 E^2$ represents energy density in an electric field. Energy density has the dimensions of energy per unit volume.

1. Permittivity of Free Space ($\varepsilon_0$):

   The permittivity of free space, also known as the electric constant, has dimensions of $[M^{-1}L^{-3}T^4A^2]$, where A is the dimension of current.

2. Electric Field ($E$):

   The electric field has dimensions of $[MLT^{-3}A^{-1}]$.

3. Combining the Dimensions:

   We need to find the dimensions for $\varepsilon_0 E^2$.

   So, $\varepsilon_0 E^2$ will have dimensions:

   • $[M^{-1}L^{-3}T^4A^2] \cdot ( [MLT^{-3}A^{-1}] )^2$
   • Calculating, we get: $[M^{-1}L^{-3}T^4A^2] \cdot [M^2L^2T^{-6}A^{-2}]$
   • Combine the powers: $[M^{-1+2}L^{-3+2}T^{4-6}A^{2-2}]$ = $[ML^{-1}T^{-2}]$
4. Final Result:

   Thus, the dimension of the expression $\frac{1}{2}\varepsilon _0 E^2$ is $[ML^{-1}T^{-2}]$, which corresponds to energy per unit volume, as expected.

Hence, the correct option is $ML^{-1}T^{-2}$.