GSO001 Problem 18

From Gauss's law in differential form:

$$\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}$$∇⋅E=ε0​ρ​

Computing the divergence of the given electric field:

$$\nabla \cdot \vec{E} = \frac{\partial}{\partial x}(k x y^2) + \frac{\partial}{\partial y}(k x^2 y) + \frac{\partial}{\partial z}(\gamma z^3)$$∇⋅E=∂x∂​(kxy2)+∂y∂​(kx2y)+∂z∂​(γz3) $$\nabla \cdot \vec{E} = k y^2 + k x^2 + 3\gamma z^2 = k(x^2 + y^2) + 3\gamma z^2$$∇⋅E=ky2+kx2+3γz2=k(x2+y2)+3γz2

The total charge in region $V$V:

$$Q = \iiint_V \rho \,dV = \varepsilon_0 \iiint_V (\nabla \cdot \vec{E}) \,dV$$Q=∭V​ρdV=ε0​∭V​(∇⋅E)dV

Using cylindrical coordinates $(r, \theta, z)$(r,θ,z) with $x^2 + y^2 = r^2$x2+y2=r2 and $dV = r \,dr \,d\theta \,dz$dV=rdrdθdz:

$$Q = \varepsilon_0 \int_0^{2\pi} d\theta \int_0^R dr \int_0^H dz \left( (k r^2 + 3\gamma z^2) r \right)$$Q=ε0​∫02π​dθ∫0R​dr∫0H​dz((kr2+3γz2)r)

Since the integrand is independent of $\theta$θ, the $\theta$θ integral gives $2\pi$2π:

$$Q = 2\pi \varepsilon_0 \int_0^R \left( \int_0^H (k r^3 + 3\gamma r z^2) \,dz \right) \,dr$$Q=2πε0​∫0R​(∫0H​(kr3+3γrz2)dz)dr

Integrating over $z$z:

$$\int_0^H (k r^3 + 3\gamma r z^2) \,dz = k H r^3 + \gamma H^3 r$$∫0H​(kr3+3γrz2)dz=kHr3+γH3r

Integrating over $r$r from $0$0 to $R$R:

$$Q = 2\pi \varepsilon_0 \left( \frac{1}{4} k H R^4 + \frac{1}{2} \gamma H^3 R^2 \right)$$Q=2πε0​(41​kHR4+21​γH3R2)

Substituting $k = 2$k=2, $\gamma = 1$γ=1, $R = 3$R=3, $H = 4$H=4:

• First term: $\frac{1}{4} \times 2 \times 4 \times 3^4 = 2 \times 81 = 162$41​×2×4×34=2×81=162
• Second term: $\frac{1}{2} \times 1 \times 4^3 \times 3^2 = \frac{1}{2} \times 64 \times 9 = 288$21​×1×43×32=21​×64×9=288

Sum: $162 + 288 = 450$162+288=450

$$Q = 2\pi \varepsilon_0 \times 450 = 900\pi \varepsilon_0$$Q=2πε0​×450=900πε0​

Substituting $\varepsilon_0 = \frac{1}{36\pi \times 10^9}\text{ F/m}$ε0​=36π×1091​ F/m:

$$Q = 900\pi \times \frac{1}{36\pi \times 10^9} = \frac{900}{36} \times 10^{-9} = 25 \times 10^{-9}\text{ C}$$Q=900π×36π×1091​=36900​×10−9=25×10−9 C

Since $1\text{ nC} = 10^{-9}\text{ C}$1 nC=10−9 C, the total charge is $25\text{ nC}$25 nC.

Therefore, the answer is $25$25.