The electric potential $$V(x)$$ in a region around the origin is given by $$V(x) = 4x^2$$ volts. The electric charge enclosed in a cube of 1 m side with its centre at the origin is (in coulomb)

Solution & Explanation

We are given the electric potential distribution $$V(x)$$ in a region around the origin as a function of the $$x$$-coordinate:

$$V(x) = 4x^2 \,\, \text{volts}$$

The electric field vector ($$\vec{E}$$) is related to the spatial gradient of the electric potential by the relation $$\vec{E} = -\vec{\nabla}V$$. Since the potential varies only along the $$x$$-axis, the field components along the $$y$$ and $$z$$ directions are zero:

$$E_x = -\frac{dV}{dx} = -\frac{d}{dx}(4x^2) = -8x \,\, \text{V/m}$$

$$E_y = 0, \quad E_z = 0$$

Thus, the electric field vector is given by: $$\vec{E} = -8x\,\hat{i}$$

Consider a cube of side length $$a = 1 \,\, \text{m}$$ centered at the origin. The boundaries of the cube along the $$x$$-axis are located at:

$$x = -\frac{a}{2} = -0.5 \,\, \text{m} \quad \text{and} \quad x = +\frac{a}{2} = +0.5 \,\, \text{m}$$

Since the electric field has only an $$x$$-component, the net electric flux ($$\Phi$$) through the top, bottom, front, and back faces of the cube (which are parallel to the $$x$$-axis) is identically zero. Flux only passes through the two faces perpendicular to the $$x$$-axis:

• At the left face ($$x = -0.5 \,\, \text{m}$$): The outward normal vector points in the negative direction ($$\hat{n} = -\hat{i}$$).

  $$\vec{E}_{\text{left}} = -8(-0.5)\,\hat{i} = +4\,\hat{i} \,\, \text{V/m}$$

  $$\Phi_{\text{left}} = \vec{E}_{\text{left}} \cdot \vec{A}_{\text{left}} = (+4\,\hat{i}) \cdot (-A\,\hat{i}) = -4A$$

• At the right face ($$x = +0.5 \,\, \text{m}$$): The outward normal vector points in the positive direction ($$\hat{n} = +\hat{i}$$).

  $$\vec{E}_{\text{right}} = -8(+0.5)\,\hat{i} = -4\,\hat{i} \,\, \text{V/m}$$

  $$\Phi_{\text{right}} = \vec{E}_{\text{right}} \cdot \vec{A}_{\text{right}} = (-4\,\hat{i}) \cdot (+A\,\hat{i}) = -4A$$

Summing the contributions over all surfaces gives the total net electric flux escaping the closed cube volume:

$$\Phi_{\text{net}} = \Phi_{\text{left}} + \Phi_{\text{right}} = -4A - 4A = -8A$$

Since the cross-sectional area of a face of the cube is $$A = \text{side}^2 = 1^2 = 1 \,\, \text{m}^2$$, we find:

$$\Phi_{\text{net}} = -8(1) = -8 \,\, \text{V}\cdot\text{m}$$

According to Gauss's Law, the net electric flux through any closed surface is proportional to the total enclosed charge ($$Q_{\text{enclosed}}$$) divided by the permittivity of free space ($$\varepsilon_0$$):

$$\Phi_{\text{net}} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$$

$$-8 = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \implies Q_{\text{enclosed}} = -8\varepsilon_0 \,\, \text{C}$$

Concept Check: Because the potential function curves upward quadratically away from the origin, the resulting electric field lines point inward toward the center from both sides. This net inward convergence represents a negative total flux, confirming that a net negative charge density of exactly $$-8\varepsilon_0$$ is enclosed within the cube boundaries.

Correct Option Key: Option D ($$-8\varepsilon_0$$)