Easy MCQ +4 / -1 PYQ · JEE Mains 2023

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 :

  1. A $\mathrm{Nm}^{2} \mathrm{C} ; \mathrm{Nm}^{3} \mathrm{C}$
  2. B $\mathrm{Nm}^{3} \mathrm{C}^{-1} ; \mathrm{Nm}^{2} \mathrm{C}^{-1}$
  3. C $\mathrm{Nm}^{3} \mathrm{C} ; \mathrm{Nm}^{2} \mathrm{C}$
  4. D $\mathrm{Nm}^{2} \mathrm{C}^{-1} ; \mathrm{Nm}^{3} \mathrm{C}^{-1}$ Correct answer

Solution

<p>$$\overrightarrow E = \left( {{A \over {{x^2}}}\widehat i + {B \over {{y^3}}}\widehat j} \right)$$</p> <p>$$\left[ {{A \over {{x^2}}}} \right] = [E] = \left[ {{F \over q}} \right] = \left[ {{N \over C}} \right] = N{C^{ - 1}}$$</p> <p>$\mathrm{[A] = (N{m^2}{C^{ - 1}})}$</p> <p>$\mathrm{[B] = N{m^3}{C^{ - 1}}}$</p>

About this question

Subject: Physics · Chapter: Electrostatics · Topic: Electric Field and Field Lines

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