The electric field in a region is given by $\vec E =\frac{E_0×\hat i}{b}$. Find the charge contained in the cubical volume bounded by the $x = 0, x = a, y = 0, y = a$ and $z = 0, z= a$. Take $E_0 = 6 × 10^3 N/C$, a = 1 cm and b = 2 cm

$2.65 × 10^{-12} C$

The correct answer is Option (4) → $2.65 × 10^{-12} C$

Assume $\displaystyle \vec{E}=\frac{E_0 x}{b}\,\hat{i}$ (so that $\vec{E}$ varies with $x$).

Divergence: $\displaystyle \nabla\cdot\vec{E}=\frac{\partial E_x}{\partial x}=\frac{E_0}{b}$.

Charge density: $\displaystyle \rho=\varepsilon_0(\nabla\cdot\vec{E})=\varepsilon_0\frac{E_0}{b}$.

Total charge in the cube: $\displaystyle Q=\rho \, a^3=\varepsilon_0\frac{E_0}{b}\,a^3$.

Substitute $\varepsilon_0=8.85\times10^{-12}\,\text{F/m}$, $E_0=6.0\times10^{3}\,\text{N/C}$, $b=2\,\text{cm}=0.02\,\text{m}$, $a=1\,\text{cm}=0.01\,\text{m}$:

$\displaystyle Q=8.85\times10^{-12}\cdot\frac{6.0\times10^{3}}{0.02}\cdot(0.01)^3 =2.655\times10^{-12}\ \text{C}$

Answer: $Q \approx 2.66\times10^{-12}\ \text{C}$