Question

The electric potential, $V$ in a certain region of space depends only on the $x$ coordinate as $V = -a_0x^4 + V_0$, where $a_0$ and $V_0$ are constants. Consider a cuboid in space of dimensions $L \times \frac{L}{2} \times \frac{L}{2}$ whose length stretches along the $x$-axis from $x = 0$ to $x = L$. Then choose the correct option(s).

• A. The charge contained within the cuboid is $\epsilon_0 a_0 L^5$
• B. The charge contained within the cuboid is $\frac{\epsilon_0 a_0 L^5}{2}$
• C. The electric flux through the surface of the cuboid is $\frac{a_0 L^5}{2}$
• D. The charge density at $x = \frac{L}{2}$ is $3\epsilon_0 a_0 L^2$

Answer 1

Answer: (A), (D)

(A), (D)

Answer 2

The electric potential in a certain region of space is given by $V(x) = -a_0x^4 + V_0$.

The electric field $\vec{E}$ is related to the potential by $\vec{E} = -\nabla V$. Since $V$ depends only on $x$, the electric field is in the $x$-direction:

$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(-a_0x^4 + V_0) = -(-4a_0x^3) = 4a_0x^3$.

So, the electric field is $\vec{E}(x) = 4a_0x^3 \hat{i}$.

The charge density $\rho$ is related to the electric field by Gauss's law in differential form: $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$.

$\rho = \epsilon_0 (\nabla \cdot \vec{E})$.

Since $\vec{E} = 4a_0x^3 \hat{i}$, the divergence is:

$\nabla \cdot \vec{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = \frac{\partial}{\partial x}(4a_0x^3) + 0 + 0 = 12a_0x^2$.

So, the charge density is $\rho(x) = \epsilon_0 (12a_0x^2) = 12\epsilon_0 a_0 x^2$.

Let's check the given options:

(D) The charge density at $x = \frac{L}{2}$ is $3\epsilon_0 a_0 L^2$.

Substitute $x = \frac{L}{2}$ into the expression for $\rho(x)$:

$\rho(\frac{L}{2}) = 12\epsilon_0 a_0 (\frac{L}{2})^2 = 12\epsilon_0 a_0 \frac{L^2}{4} = 3\epsilon_0 a_0 L^2$.

Option (D) is correct.

The cuboid has dimensions $L \times \frac{L}{2} \times \frac{L}{2}$ and stretches along the $x$-axis from $x = 0$ to $x = L$. Let the cuboid occupy the region $0 \le x \le L$, $0 \le y \le L/2$, $0 \le z \le L/2$.

The charge contained within the cuboid is $Q = \int_V \rho \, dV$.

$Q = \int_0^{L/2} \int_0^{L/2} \int_0^L \rho(x) \, dx \, dy \, dz$

$Q = \int_0^{L/2} dy \int_0^{L/2} dz \int_0^L (12\epsilon_0 a_0 x^2) \, dx$

$Q = (\frac{L}{2})(\frac{L}{2}) \int_0^L 12\epsilon_0 a_0 x^2 \, dx$

$Q = \frac{L^2}{4} \cdot 12\epsilon_0 a_0 [\frac{x^3}{3}]_0^L$

$Q = 3\epsilon_0 a_0 L^2 (\frac{L^3}{3} - 0) = \epsilon_0 a_0 L^5$.

(A) The charge contained within the cuboid is $\epsilon_0 a_0 L^5$.

This matches our calculation. Option (A) is correct.

(B) The charge contained within the cuboid is $\frac{\epsilon_0 a_0 L^5}{2}$.

This does not match our calculation. Option (B) is incorrect.

The electric flux through the surface of the cuboid can be calculated using Gauss's law: $\Phi_E = \frac{Q_{enclosed}}{\epsilon_0}$.

Using the charge enclosed we calculated:

$\Phi_E = \frac{\epsilon_0 a_0 L^5}{\epsilon_0} = a_0 L^5$.

Alternatively, we can calculate the flux by integrating $\vec{E} \cdot d\vec{A}$ over the surface of the cuboid.

The cuboid has 6 faces. The electric field is $\vec{E}(x) = 4a_0x^3 \hat{i}$.

The faces perpendicular to the $x$-axis are at $x=0$ and $x=L$.

• Face at $x=0$: Area vector $d\vec{A} = -dy dz \hat{i}$. $\vec{E}(0) = 4a_0(0)^3 \hat{i} = 0$. Flux $\Phi_1 = \int \vec{E}(0) \cdot d\vec{A} = 0$.

• Face at $x=L$: Area vector $d\vec{A} = dy dz \hat{i}$. $\vec{E}(L) = 4a_0L^3 \hat{i}$. Flux $\Phi_2 = \int_{0}^{L/2} \int_{0}^{L/2} (4a_0L^3 \hat{i}) \cdot (dy dz \hat{i}) = 4a_0L^3 \int_0^{L/2} dy \int_0^{L/2} dz = 4a_0L^3 (\frac{L}{2})(\frac{L}{2}) = a_0L^5$.

The faces perpendicular to the $y$ and $z$ axes have area vectors in the $\hat{j}$ or $\hat{k}$ directions. Since $\vec{E}$ is only in the $\hat{i}$ direction, the dot product $\vec{E} \cdot d\vec{A}$ is zero for these faces.

Total flux $\Phi_E = \Phi_1 + \Phi_2 + \Phi_{y faces} + \Phi_{z faces} = 0 + a_0L^5 + 0 + 0 = a_0L^5$.

(C) The electric flux through the surface of the cuboid is $\frac{a_0 L^5}{2}$.

This does not match our calculation. Option (C) is incorrect.

The correct options are (A) and (D).