Question

A cuboid with dimension a = 0.5 m, b = 1 m and c = 2 m is placed symmetrically about origin in x-z plane in non-uniform electric field $\vec{E} = (3 + 2x^2) \hat{i}$ N/C. The net flux through the cuboid is ________ N. m²/C

Answer 1

0 N.m²/C

Answer 2

The electric field is given by $\vec{E} = (3 + 2x^2) \hat{i}$ N/C. This means the electric field is directed only along the x-axis, and its magnitude depends on the x-coordinate.

The cuboid has dimensions a = 0.5 m, b = 1 m, and c = 2 m. It is placed symmetrically about the origin. This implies:

• The extent along the x-axis is from $x = -a/2$ to $x = +a/2$. So, $x = -0.5/2 = -0.25$ m to $x = +0.5/2 = +0.25$ m.
• The extent along the y-axis is from $y = -b/2$ to $y = +b/2$. So, $y = -1/2 = -0.5$ m to $y = +1/2 = +0.5$ m.
• The extent along the z-axis is from $z = -c/2$ to $z = +c/2$. So, $z = -2/2 = -1$ m to $z = +1$ m.

The electric flux through a closed surface is given by $\Phi = \oint \vec{E} \cdot d\vec{A}$. We need to calculate the flux through each of the six faces of the cuboid.

1. Faces perpendicular to the y-axis (y = -0.5 m and y = +0.5 m):
   For these faces, the area vector $d\vec{A}$ is in the $\pm \hat{j}$ direction. Since the electric field $\vec{E}$ is in the $\hat{i}$ direction, their dot product $\vec{E} \cdot d\vec{A}$ will be zero ($\hat{i} \cdot \hat{j} = 0$). Therefore, the flux through these faces is zero.

2. Faces perpendicular to the z-axis (z = -1 m and z = +1 m):
   For these faces, the area vector $d\vec{A}$ is in the $\pm \hat{k}$ direction. Since the electric field $\vec{E}$ is in the $\hat{i}$ direction, their dot product $\vec{E} \cdot d\vec{A}$ will be zero ($\hat{i} \cdot \hat{k} = 0$). Therefore, the flux through these faces is zero.

3. Faces perpendicular to the x-axis:
   These are the only faces that will contribute to the net flux. The area of these faces is $A_x = b \times c = 1 \text{ m} \times 2 \text{ m} = 2 \text{ m}^2$.

   • Left face (at x = -0.25 m):
     The electric field at this face is $\vec{E}_{left} = (3 + 2(-0.25)^2) \hat{i}$.
     $(-0.25)^2 = (1/4)^2 = 1/16 = 0.0625$.
     $\vec{E}_{left} = (3 + 2(0.0625)) \hat{i} = (3 + 0.125) \hat{i} = 3.125 \hat{i}$ N/C.
     The area vector for the left face is $\vec{A}_{left} = A_x (-\hat{i}) = 2 (-\hat{i})$ m$^2$. (The outward normal points in the $-\hat{i}$ direction).
     The flux through the left face is $\Phi_{left} = \vec{E}_{left} \cdot \vec{A}_{left} = (3.125 \hat{i}) \cdot (2 (-\hat{i})) = -3.125 \times 2 = -6.25$ N.m²/C.

   • Right face (at x = +0.25 m):
     The electric field at this face is $\vec{E}_{right} = (3 + 2(+0.25)^2) \hat{i}$.
     $\vec{E}_{right} = (3 + 2(0.0625)) \hat{i} = (3 + 0.125) \hat{i} = 3.125 \hat{i}$ N/C.
     The area vector for the right face is $\vec{A}_{right} = A_x (+\hat{i}) = 2 (+\hat{i})$ m$^2$. (The outward normal points in the $+\hat{i}$ direction).
     The flux through the right face is $\Phi_{right} = \vec{E}_{right} \cdot \vec{A}_{right} = (3.125 \hat{i}) \cdot (2 (+\hat{i})) = +3.125 \times 2 = +6.25$ N.m²/C.

Net Flux through the Cuboid:
The total net flux is the sum of the fluxes through all six faces:
$\Phi_{net} = \Phi_{left} + \Phi_{right} + \Phi_{y_{faces}} + \Phi_{z_{faces}}$
$\Phi_{net} = -6.25 + 6.25 + 0 + 0 = 0$ N.m²/C.

This result is expected because the electric field component along x, $E_x = (3 + 2x^2)$, is an even function of x (i.e., $E_x(-x) = E_x(x)$). When a closed surface is placed symmetrically about the origin along the direction of an electric field that is an even function of the position coordinate, the incoming flux on one side is exactly cancelled by the outgoing flux on the opposite side.