JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 10)
Electric field in a certain region is given by $$\overrightarrow{\mathrm{E}}=\left(\frac{\mathrm{A}}{x^{2}} \hat{i}+\frac{\mathrm{B}}{y^{3}} \hat{j}\right) \text {. The } \mathrm{SI} \text { unit of } \mathrm{A} \text { and } \mathrm{B}$$ are :
$$\mathrm{Nm}^{2} \mathrm{C} ; \mathrm{Nm}^{3} \mathrm{C}$$
$$\mathrm{Nm}^{3} \mathrm{C}^{-1} ; \mathrm{Nm}^{2} \mathrm{C}^{-1}$$
$$\mathrm{Nm}^{3} \mathrm{C} ; \mathrm{Nm}^{2} \mathrm{C}$$
$$\mathrm{Nm}^{2} \mathrm{C}^{-1} ; \mathrm{Nm}^{3} \mathrm{C}^{-1}$$
Explanation
$$\overrightarrow E = \left( {{A \over {{x^2}}}\widehat i + {B \over {{y^3}}}\widehat j} \right)$$
$$\left[ {{A \over {{x^2}}}} \right] = [E] = \left[ {{F \over q}} \right] = \left[ {{N \over C}} \right] = N{C^{ - 1}}$$
$$\mathrm{[A] = (N{m^2}{C^{ - 1}})}$$
$$\mathrm{[B] = N{m^3}{C^{ - 1}}}$$
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