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Step-by-Step Solution
Step 1: Identify the given electric field expression
The electric field is given by
$ \displaystyle \overrightarrow{E} \;=\;\frac{A}{x^2}\,\hat{i} \;+\;\frac{B}{y^3}\,\hat{j}. $
We need to find the SI units of the constants $A$ and $B$.
Step 2: Recall the SI unit of electric field
In SI units, the electric field $\overrightarrow{E}$ has the unit $ \mathrm{N\,C^{-1}} $, which is equivalent to $ \mathrm{V\,m^{-1}} $ but commonly expressed as $ \mathrm{N\,C^{-1}} $.
Step 3: Determine the unit requirements for $A$
The term involving $A$ is $ \frac{A}{x^2} $. Since this expression must have the same dimensions as the electric field $ \mathrm{N\,C^{-1}} $, we write:
\[
\left[\frac{A}{x^2}\right] \;=\; \left[\mathrm{N\,C^{-1}}\right].
\]
Hence,
\[
[A] \;=\; \left[\mathrm{N\,C^{-1}}\right] \times \left[x^2\right].
\]
Because $x$ is a length with SI unit $\mathrm{m}$, then $x^2$ has unit $\mathrm{m^2}$. Therefore,
\[
[A] \;=\; \mathrm{N\,C^{-1}} \times \mathrm{m^2}.
\]
Since $\mathrm{N} = \mathrm{kg\,m\,s^{-2}}$, we get
\[
[A] \;=\; \mathrm{kg\,m\,s^{-2} \,C^{-1}} \times \mathrm{m^2}
\;=\; \mathrm{kg\,m^3\,s^{-2}\,C^{-1}} \;=\; \mathrm{N\,m^2\,C^{-1}}.
\]
Step 4: Determine the unit requirements for $B$
Similarly, the term involving $B$ is $ \frac{B}{y^3} $. Following the same reasoning,
\[
\left[\frac{B}{y^3}\right] \;=\; \left[\mathrm{N\,C^{-1}}\right].
\]
Thus,
\[
[B] \;=\; \left[\mathrm{N\,C^{-1}}\right] \times \left[y^3\right].
\]
Since $y$ is also a length with SI unit $\mathrm{m}$, $y^3$ has unit $\mathrm{m^3}$. Therefore,
\[
[B] \;=\; \mathrm{N\,C^{-1}} \times \mathrm{m^3}
\;=\; \mathrm{N\,m^3\,C^{-1}}.
\]
Step 5: State the final units
Hence, the SI units of $A$ and $B$ are:
\[
[A] = \mathrm{N\,m^2\,C^{-1}}, \quad
[B] = \mathrm{N\,m^3\,C^{-1}}.
\]
