The electric field is given by $\vec{E}=\frac{A}{x^{3}} \hat{i}+B y \hat{j}+C z^{2} \hat{k} .$ The SI units…
$\vec{E}=\frac{A}{x^{3}} \hat{i}+B y \hat{j}+C z^{2} \hat{k} .$ The SI units of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$
are respectively: [where $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are in $\mathrm{m}]$
- $\frac{N-m^{3}}{C}, V / m^{2}, \mathrm{~N} / \mathrm{m}^{2}-\mathrm{C}$
- $\mathrm{V}-\mathrm{m}^{2}, \mathrm{~V} / \mathrm{m}, \mathrm{N} / \mathrm{m}^{2}-\mathrm{C}$
- $\mathrm{V} / \mathrm{m}^{2}, \mathrm{~V} / \mathrm{m}, \mathrm{N}-\mathrm{C} / \mathrm{m}^{2}$
- $\mathrm{V} / \mathrm{m}, \mathrm{N}-\mathrm{m}^{3} / \mathrm{C}, \mathrm{N}-\mathrm{C} / \mathrm{m}$
Solution
So, unit of $\mathrm{A}=\mathrm{E} \times \mathrm{x}^{3}=\frac{\mathrm{N}-\mathrm{m}^{3}}{\mathrm{C}}=\left(\mathrm{V}-\mathrm{m}^{2}\right)$
Unit of $\mathrm{B}=\mathrm{E} / \mathrm{y}=\frac{N}{m c}=\left(\frac{V}{m^{2}}\right)$
Unit of $\mathrm{C}=\mathrm{E} / \mathrm{z}^{2}=\frac{N}{m^{2} c}=\left(\frac{V}{m^{3}}\right)$ .
Asked in: JEE Mains - Units and Dimensions - Test 1