Question

A cube of side a has one corner at the origin and its sides lie along the x, y, and z axes, respectively. There is a field given by $\hat{E}=b(1+x)\hat{i}$. The net charge enclosed by the cube is $na^3b\epsilon_0$ where n is________.

Answer 1

1

Answer 2

The electric field is given by $\vec{E} = b(1+x)\hat{i}$.

The cube has sides of length $a$ and is located with one corner at the origin and sides along the x, y, and z axes. The faces of the cube are at $x=0, x=a, y=0, y=a, z=0, z=a$.

We use Gauss's Law, which states that the total electric flux through a closed surface is equal to the net charge enclosed divided by the permittivity of free space ($\epsilon_0$):

$\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$

The total flux through the cube is the sum of the fluxes through its six faces.

The electric field is in the x-direction. For the faces perpendicular to the y-axis (y=0 and y=a) and the z-axis (z=0 and z=a), the area vectors are perpendicular to the electric field ($\hat{j}$ or $\hat{k}$ are perpendicular to $\hat{i}$). Thus, the flux through these four faces is zero.

We only need to calculate the flux through the faces perpendicular to the x-axis.

1. Face at $x=0$:

The area vector for this face points in the negative x-direction, $\vec{A}_{x=0} = -a^2\hat{i}$.

The electric field at this face is $\vec{E}(x=0) = b(1+0)\hat{i} = b\hat{i}$.

The flux through this face is $\Phi_{x=0} = \vec{E}(x=0) \cdot \vec{A}_{x=0} = (b\hat{i}) \cdot (-a^2\hat{i}) = -ba^2$.

2. Face at $x=a$:

The area vector for this face points in the positive x-direction, $\vec{A}_{x=a} = a^2\hat{i}$.

The electric field at this face is $\vec{E}(x=a) = b(1+a)\hat{i}$.

The flux through this face is $\Phi_{x=a} = \vec{E}(x=a) \cdot \vec{A}_{x=a} = (b(1+a)\hat{i}) \cdot (a^2\hat{i}) = b(1+a)a^2 = ba^2 + ba^3$.

The total flux through the cube is the sum of the fluxes through all six faces:

$\Phi_{net} = \Phi_{x=0} + \Phi_{x=a} + \Phi_{y=0} + \Phi_{y=a} + \Phi_{z=0} + \Phi_{z=a}$

$\Phi_{net} = -ba^2 + (ba^2 + ba^3) + 0 + 0 + 0 + 0$

$\Phi_{net} = -ba^2 + ba^2 + ba^3 = ba^3$.

According to Gauss's Law, the net charge enclosed by the cube is:

$q_{enclosed} = \epsilon_0 \Phi_{net} = \epsilon_0 (ba^3)$.

The problem states that the net charge enclosed by the cube is $na^3b\epsilon_0$.

Comparing our result with the given form:

$q_{enclosed} = ba^3\epsilon_0$

$q_{enclosed} = na^3b\epsilon_0$

Equating the two expressions for $q_{enclosed}$:

$ba^3\epsilon_0 = na^3b\epsilon_0$

Assuming $a \ne 0$, $b \ne 0$, and $\epsilon_0 \ne 0$, we can cancel out the common terms $a^3$, $b$, and $\epsilon_0$ from both sides:

$1 = n$

Thus, the value of $n$ is 1.