Electrostatics

JEE Main level, numerical answer, medium

292 of 311

Question 292 of 311

The total charge enclosed in an incremental volume of 2×10−9m32\times {10}^{-9}{m}^{3} located at the origin is nC,nC, if electric flux density of its field is found asD=e−xsin⁡yi^−e−xcosyj^+2zk^Cm−2D={e}^{-x}\sin y\hat{i}-{e}^{-x}cosy\hat{j}+2z\hat{k}C{m}^{-2}

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Answer: 4

Electric flux density,
(D⃗)=chargeArea×r^=Q4πr2r^=ϵ0(Q4πϵ0r2r^)\left(\vec{D}\right)=\dfrac{\text{charge}}{\text{Area}}\times \hat{r}=\dfrac{Q}{4\pi {r}^{2}}\hat{r}={ϵ}_{0}\left(\dfrac{Q}{4\pi {ϵ}_{0}{r}^{2}}\hat{r}\right)
⇒E⃗=D⃗ϵ0=e−xsinyi^−e−xcos(j^+2zk)^ϵ0\Rightarrow \vec{E}=\dfrac{\vec{D}}{{ϵ}_{0}}=\dfrac{{e}^{-x}siny\hat{i}-{e}^{-x}cos(\hat{j}+2z\hat{k)}}{{ϵ}_{0}}
Also, by Gauss's law,
ρϵ0=(∂∂xi^+∂∂yj^+∂∂zk^)⋅E⃗=(∂∂xi^+∂∂yj^+∂∂zk^)⋅D⃗ϵ0\dfrac{\rho }{{ϵ}_{0}}=\left(\dfrac{\partial }{\partial x}\hat{i}+\dfrac{\partial }{\partial y}\hat{j}+\dfrac{\partial }{\partial z}\hat{k}\right)\cdot \vec{E}=\left(\dfrac{\partial }{\partial x}\hat{i}+\dfrac{\partial }{\partial y}\hat{j}+\dfrac{\partial }{\partial z}\hat{k}\right)\cdot \dfrac{\vec{D}}{{ϵ}_{0}}
⇒ρ=∂∂x(e−xsin⁡y)+∂∂y(−e−xcos⁡y)+∂∂z(2z)\Rightarrow \rho =\dfrac{\partial }{\partial x}\left({e}^{-x}\sin y\right)+\dfrac{\partial }{\partial y}\left(-{e}^{-x}\cos y\right)+\dfrac{\partial }{\partial z}(2z)
ρ=−e−xsin⁡y+e−xsin⁡y+2\rho =-{e}^{-x}\sin y+{e}^{-x}\sin y+2
At origin ρ=−e−0sin⁡0+e−0sin⁡0+2\rho =-{e}^{-0}\sin 0+{e}^{-0}\sin 0+2
ρ=2Cm−3\rho =2C{m}^{-3}
Charge =ρ×=\rho \times volume =2×2×10−9=4×10−9==2\times 2\times {10}^{-9}=4\times {10}^{-9}=
4nC4nC

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