Question 56: 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: