The electric field in a region is given by $\displaystyle \vec{E}=\frac{E_0 x}{b}\,\hat{i}$ . 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) → $2.65 × 10^{-12} C$

Given: $\vec{E} = \frac{E_0 x}{b}\hat{i}$

Step 1: Identify the surfaces contributing to Flux.

Since the field is only in the $\hat{i}$ direction, flux ($\phi$) through faces parallel to the x-axis is zero. We only check the faces at $x = 0$ and $x = a$.

• At $x = 0$: $E_1 = 0 ⇒\phi_1 = 0$
• At $x = a$: $E_2 = \frac{E_0 a}{b} ⇒\phi_2 = E_2 \cdot \text{Area} = \frac{E_0 a}{b} \cdot a^2 = \frac{E_0 a^3}{b}$

Step 2: Apply Gauss's Law.

The net flux is $\phi_{\text{net}} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$

$Q = \varepsilon_0 \cdot \phi_{\text{net}} = \varepsilon_0 \cdot \frac{E_0 a^3}{b}$

• $\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{Nm}^2$
• $E_0 = 6.0 \times 10^3\text{ N/C}$
• $a = 10^{-2}\text{ m},\ b = 2 \times 10^{-2}\text{ m}$

$Q = 8.85 \times 10^{-12} \cdot \frac{6.0 \times 10^3 \cdot (10^{-2})^3}{2 \times 10^{-2}} = 2.655 \times 10^{-12}\text{ C}$