Question
The electric field is given by $$\vec E = \frac{A}{{{x^3}}}\hat i + By\hat j + C{z^2}\hat k.$$ The SI units of $$A,B$$ and $$C$$ are respectively: [where $$x,y$$ and $$z$$ are in $$m$$ ]
A.
$$\frac{{N - {m^3}}}{C},\frac{V}{{{m^2}}},\frac{N}{{{m^2} - C}}$$
B.
$$V - {m^2},\frac{V}{m},\frac{N}{{{m^2} - C}}$$
C.
$$\frac{V}{{{m^2}}},\frac{V}{m},\frac{{N - C}}{{{m^2}}}$$
D.
$$\frac{V}{m},\frac{{N - {m^3}}}{C},\frac{{N - C}}{m}$$
Answer :
$$\frac{{N - {m^3}}}{C},\frac{V}{{{m^2}}},\frac{N}{{{m^2} - C}}$$
Solution :
Here $$A,B$$ and $$C$$ must have same unit as of electric field.
$$\eqalign{
& {\text{So,}}\,\frac{A}{{{x^3}}} = \frac{N}{C} \cr
& \Rightarrow \frac{A}{{{m^3}}} \equiv \frac{N}{C} \cr
& \Rightarrow A \equiv \frac{{N - {m^3}}}{C} \cr
& {\text{and}}\,By = \frac{V}{m} \cr
& \Rightarrow Bm \equiv \frac{V}{{{m^2}}} \cr
& \Rightarrow B \equiv \frac{V}{{{m^2}}} \cr
& {\text{and}}\,C{z^2} \equiv \frac{N}{C} \cr
& \Rightarrow C{m^2} \equiv \frac{N}{C} \cr
& \Rightarrow C \equiv \frac{N}{{C - {m^2}}} \cr} $$