Question

Electric field in a certain region is given by $\overrightarrow{ E }=\left(\frac{ A }{x^2} \hat{i}+\frac{ B }{y^3} \hat{j}\right)$ The $SI$ unit of $A$ and $B$ are :

• A. $Nm ^2 C ^{-1} ; Nm ^3 C ^{-1}$
• B. $Nm ^3 C ; Nm ^2 C$
• C. $Nm ^2 C ; Nm ^3 C$
• D. $Nm ^3 C ^{-1} ; Nm ^2 C ^{-1}$

Answer 1

The Correct Option is A

To determine the SI units of \( A \) and \( B \) from the given electric field vector \(\overrightarrow{E} = \left(\frac{A}{x^2} \hat{i} + \frac{B}{y^3} \hat{j}\right)\), we need to analyze the dimensional formulas involved in electrical field expressions.

The electric field \(\overrightarrow{E}\) is measured in Newtons per Coulomb (N/C) in SI units. This can also be expressed as \((kg \cdot m/s^2)/C\).

Let's start with the component \(\frac{A}{x^2}\):

• The term \(\frac{A}{x^2}\) must have the same dimensional formula as the electric field, which is \(N/C\).
• We express \(x\) as being of dimension [L], which implies [\(\text{length}\)] or meters (\(m\)).
• Thus, \(\frac{A}{x^2}\) has the dimension [A]/[L\(^2\)] = [A]/\((m^2)\).
• Therefore, A must have dimension such that [A]/[m\(^2\)] = [N/C].

This means \(A\) must have dimensions [N \cdot m^2/C] to match the units of the electric field in the direction \(\hat{i}\). So, the unit of \(A\) is \(Nm^2C^{-1}\).

Now let's consider \(\frac{B}{y^3}\):

• Similarly, \(\frac{B}{y^3}\) has the same dimensional formula as the electric field, \(N/C\).
• With \(y\) being a length (\(m\)) and the expression being \(\frac{B}{m^3}\), we derive the dimension [B]/[L\(^3\)] = [B]/(\(m^3\))
• Therefore, B must have dimension such that [B]/[m\(^3\)] = [N/C].

Thus, \(B\) must have the unit of \(Nm^3C^{-1}\).

Conclusively, the SI units for \(A\) and \(B\) are determined as follows:

• For \(A\): \(Nm^2C^{-1}\)
• For \(B\): \(Nm^3C^{-1}\)

The correct answer among the options given is: \(Nm^2C^{-1}\) ; \(Nm^3C^{-1}\)