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

(B) $\mathrm{Nm}^{2} \mathrm{C}^{-1} ; \mathrm{Nm}^{3} \mathrm{C}^{-1}$

(A) $\mathrm{Nm}^{3} \mathrm{C}^{-1} ; \mathrm{Nm}^{2} \mathrm{C}^{-1}$

(C) $\mathrm{Nm}^{3} \mathrm{C} ; \mathrm{Nm}^{2} \mathrm{C}$

(D) $\mathrm{Nm}^{2} \mathrm{C} ; \mathrm{Nm}^{3} \mathrm{C}$

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Electrostatics

Electric field in a certain region is given by

E = (\frac{A}{x ^{2}} \hat{i} + \frac{B}{y ^{3}} \hat{j})

. The SI unit of

A

and

B

are :

Answer: (B)

Every term of

E

must carry the field's unit

N C^{- 1}

. Since

\frac{A}{x ^{2}}

has this unit,

A

carries

N C^{- 1} \cdot m^{2} = N m^{2} C^{- 1}

; likewise

\frac{B}{y ^{3}}

gives

B

the unit

N C^{- 1} \cdot m^{3} = N m^{3} C^{- 1}

. Thus

A

is measured in

N m^{2} C^{- 1}

and

B

N m^{3} C^{- 1}

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option