JEE Physics 2023 Question with Solution

Given: $\mathbf{E} = \left( \frac{A}{x^2} \hat{i} + \frac{B}{y^3} \hat{j} \right)$

Find: The SI units of $A$ and $B$.

The electric field has unit $N C^{-1}$.

From the $\hat{i}$-component,

So,

From the $\hat{j}$-component,

So,

Therefore, the correct option is B.

Given: Each term of the electric field expression must have the same unit as $\mathbf{E}$.

Find: Units of $A$ and $B$.

Since position coordinates $x$ and $y$ have unit $m$, the term $\frac{A}{x^2}$ must have unit $N C^{-1}$. Hence $A$ must supply an extra factor of $m^2$, so its unit is $N m^2 C^{-1}$.

Similarly, $\frac{B}{y^3}$ must also have unit $N C^{-1}$. Therefore $B$ must supply an extra factor of $m^3$, so its unit is $N m^3 C^{-1}$.

Therefore, the correct option is B. Note that the solution labels the option as A, but the derived units match option B in the given list.