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Question: Ex. An electric field in a region is given by $\overrightarrow{E} = x^2\hat{i}$. Find the flux passi...

Ex. An electric field in a region is given by E→=x2i^\overrightarrow{E} = x^2\hat{i}. Find the flux passing through a cube of side a. (Assume a cube of side a is placed with one corner at origin such that its sides represents x, y and z axes)

A

a4a^4

B

2a33\frac{2a^3}{3}

C

a22\frac{a^2}{2}

D

a3a^3

Answer

A

Explanation

Solution

The electric field in the region is given by E→=x2i^\overrightarrow{E} = x^2\hat{i}. The cube of side 'a' is placed with one corner at the origin, and its sides are along the x, y, and z axes. The cube is bounded by the planes x=0, x=a, y=0, y=a, z=0, z=a. The electric flux through a closed surface is given by Φ=∮E→⋅dA→\Phi = \oint \overrightarrow{E} \cdot d\overrightarrow{A}. We need to calculate the flux through each of the six faces of the cube and sum them up.

  1. Face at x=0: This face is in the yz-plane at x=0. The area vector is dA→=dA(−i^)d\overrightarrow{A} = dA (-\hat{i}). At this face, x=0, so the electric field is E→(x=0)=(0)2i^=0→\overrightarrow{E}(x=0) = (0)^2\hat{i} = \overrightarrow{0}. The flux through this face is Φ1=∫E→(x=0)⋅dA→=∫0→⋅dA→=0\Phi_1 = \int \overrightarrow{E}(x=0) \cdot d\overrightarrow{A} = \int \overrightarrow{0} \cdot d\overrightarrow{A} = 0.

  2. Face at x=a: This face is a square in the plane x=a. The area vector is dA→=dA(+i^)d\overrightarrow{A} = dA (+\hat{i}). At this face, x=a, so the electric field is E→(x=a)=a2i^\overrightarrow{E}(x=a) = a^2\hat{i}. The flux through this face is Φ2=∫E→(x=a)⋅dA→=∫(a2i^)⋅(dAi^)=∫a2dA\Phi_2 = \int \overrightarrow{E}(x=a) \cdot d\overrightarrow{A} = \int (a^2\hat{i}) \cdot (dA \hat{i}) = \int a^2 dA. Since the face is a square of side 'a', its area is A=a2A = a^2. Φ2=a2∫dA=a2(a2)=a4\Phi_2 = a^2 \int dA = a^2 (a^2) = a^4.

  3. Face at y=0: This face is in the xz-plane at y=0. The area vector is dA→=dA(−j^)d\overrightarrow{A} = dA (-\hat{j}). The electric field is E→=x2i^\overrightarrow{E} = x^2\hat{i}. The flux through this face is Φ3=∫E→⋅dA→=∫(x2i^)⋅(dA(−j^))=0\Phi_3 = \int \overrightarrow{E} \cdot d\overrightarrow{A} = \int (x^2\hat{i}) \cdot (dA (-\hat{j})) = 0, since i^⋅j^=0\hat{i} \cdot \hat{j} = 0.

  4. Face at y=a: This face is in the xz-plane at y=a. The area vector is dA→=dA(+j^)d\overrightarrow{A} = dA (+\hat{j}). The electric field is E→=x2i^\overrightarrow{E} = x^2\hat{i}. The flux through this face is Φ4=∫E→⋅dA→=∫(x2i^)⋅(dA(+j^))=0\Phi_4 = \int \overrightarrow{E} \cdot d\overrightarrow{A} = \int (x^2\hat{i}) \cdot (dA (+\hat{j})) = 0, since i^⋅j^=0\hat{i} \cdot \hat{j} = 0.

  5. Face at z=0: This face is in the xy-plane at z=0. The area vector is dA→=dA(−k^)d\overrightarrow{A} = dA (-\hat{k}). The electric field is E→=x2i^\overrightarrow{E} = x^2\hat{i}. The flux through this face is Φ5=∫E→⋅dA→=∫(x2i^)⋅(dA(−k^))=0\Phi_5 = \int \overrightarrow{E} \cdot d\overrightarrow{A} = \int (x^2\hat{i}) \cdot (dA (-\hat{k})) = 0, since i^⋅k^=0\hat{i} \cdot \hat{k} = 0.

  6. Face at z=a: This face is in the xy-plane at z=a. The area vector is dA→=dA(+k^)d\overrightarrow{A} = dA (+\hat{k}). The electric field is E→=x2i^\overrightarrow{E} = x^2\hat{i}. The flux through this face is Φ6=∫E→⋅dA→=∫(x2i^)⋅(dA(+k^))=0\Phi_6 = \int \overrightarrow{E} \cdot d\overrightarrow{A} = \int (x^2\hat{i}) \cdot (dA (+\hat{k})) = 0, since i^⋅k^=0\hat{i} \cdot \hat{k} = 0.

The total flux passing through the cube is the sum of the fluxes through all six faces: Φtotal=Φ1+Φ2+Φ3+Φ4+Φ5+Φ6=0+a4+0+0+0+0=a4\Phi_{total} = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6 = 0 + a^4 + 0 + 0 + 0 + 0 = a^4.

The final answer is a4a^4.

Explanation of the solution: The electric field is E→=x2i^\overrightarrow{E} = x^2\hat{i}. The cube is from x=0 to a, y=0 to a, z=0 to a. Flux through the face at x=0: E→=0\overrightarrow{E}=0, so flux = 0. Flux through the face at x=a: E→=a2i^\overrightarrow{E}=a^2\hat{i}. Area vector is a2i^a^2\hat{i}. Flux = (a2i^)⋅(a2i^)=a4(a^2\hat{i}) \cdot (a^2\hat{i}) = a^4. Flux through faces perpendicular to y and z axes: E→\overrightarrow{E} is in i^\hat{i} direction, area vectors are in j^\hat{j} or k^\hat{k} directions. Dot product is zero, so flux = 0 for these four faces. Total flux = 0+a4+0+0+0+0=a40 + a^4 + 0 + 0 + 0 + 0 = a^4.