The total charge enclosed in an incremental volume of $2 \times 10^{-9} \mathrm{~m}^{3}$ located at the origin is ______ nC, if electric flux density of its field is found as $\mathrm{D}=\mathrm{e}^{-\mathrm{x}} \sin \mathrm{y} \hat{\mathrm{i}}-\mathrm{e}^{-\mathrm{x}} \cos \mathrm{y} \hat{\mathrm{j}}+2 \mathrm{z} \hat{\mathrm{k}} \mathrm{C} / \mathrm{m}^{2}$.

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Electrostatics

The total charge enclosed in an incremental volume of

2 \times 1 0^{- 9} m^{3}

located at the origin is ______ nC, if electric flux density of its field is found as

D = e^{- x} sin y \hat{i} - e^{- x} cos y \hat{j} + 2 z \hat{k} C / m^{2}

Answer: 4

Using Gauss's law in differential form:

\nabla \cdot D = ρ_{v}

(volume charge density). Calculate the divergence:

\nabla \cdot D = \frac{\partial}{\partial x} (e^{- x} sin y) + \frac{\partial}{\partial y} (- e^{- x} cos y) + \frac{\partial}{\partial z} (2 z) = - e^{- x} sin y + e^{- x} sin y + 2 = 2

C/m³. At the origin,

ρ_{v} = 2

C/m³. Total charge

Q = ρ_{v} \times V = 2 \times 2 \times 1 0^{- 9} = 4 \times 1 0^{- 9}

= 4

nC.

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer