Question 84: Units , Dimensions and Error Analysis

Question

If $\epsilon_{\mathrm{o}}$ is the permittivity of free space and $\mathrm{E}$ is the electric field, then $\epsilon_{\mathrm{o}} \mathrm{E}^2$ has the dimensions :

Accepted Answer

To determine the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$, we need to first understand the dimensional formulas of each component in the expression.
1.
Permittivity of free space, $\epsilon_{\mathrm{o}}$:
The permittivity of free space has the dimensions: $\left[\epsilon_{\mathrm{o}}\right] = [\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2]$.
2.
Electric field, $\mathrm{E}$:
The electric field $\mathrm{E}$ has the dimensions: $\left[\mathrm{E}\right] = [\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{A}^{-1}]$.
Now, let's calculate the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$:
$\begin{equation}  \left[\epsilon_{\mathrm{o}} \mathrm{E}^2 \right] = \left[\epsilon_{\mathrm{o}}\right] \left[\mathrm{E}\right]^2  = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left([\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{A}^{-1}]^2\right]  = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left([\mathrm{M}^2 \mathrm{L}^2 \mathrm{T}^{-6} \mathrm{A}^{-2}]\right]  \end{equation}$
Combining the dimensions:
$\begin{equation}  \left[\epsilon_{\mathrm{o}} \mathrm{E}^2 \right] = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left[\mathrm{M}^2 \mathrm{L}^2 \mathrm{T}^{-6} \mathrm{A}^{-2}\right]  = \left[\mathrm{M}^{-1 + 2} \mathrm{L}^{-3 + 2} \mathrm{T}^{4 - 6} \mathrm{A}^{2 - 2}\right]  = \left[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}\right]  \end{equation}$
Therefore, the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$ are $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$, which corresponds to Option A.
Hence, the correct answer is Option A: $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$.

Answer

$[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}]$

Answer

$[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}]$

Answer

$[\mathrm{M}^{\mathrm{0}} \mathrm{L}^{-2} \mathrm{TA}]$

Answer

$[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2]$

Units , Dimensions and Error Analysis

If \epsilon_{\mathrm{o}} is the permittivity of free space and \mathrm{E} is the electric field, then \epsilon_{\mathrm{o}} \mathrm{E}^2 has the dimensions :

Choose the correct answer: