Question

The electric field in a region is given by $ \vec{E} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3 \, N/C $. The flux of the field through a rectangular surface parallel to x-z plane is $ 6.0 \, Nm^2C^{-1} $. The area of the surface is ____ $ cm^2 $.

Hint:

The electric flux through a surface depends on the component of the electric field that is normal to the surface. When the surface is parallel to the x-z plane, its normal vector is along the y-axis, so only the y-component of the electric field contributes to the flux. Remember to convert units if the final answer requires a specific unit.

Answer 1

Correct Answer: 15

To find the area of the rectangular surface, we start with the formula for electric flux: \( \Phi = \vec{E} \cdot \vec{A} \). Here, the electric field \( \vec{E} \) is given by: \( \vec{E} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3 \, N/C \). The flux \( \Phi \) is given as \( 6.0 \, Nm^2C^{-1} \).

The surface is parallel to the x-z plane. For a surface parallel to the x-z plane, the area vector \( \vec{A} \) is perpendicular to the plane in the y-direction and can be expressed as \( \vec{A} = A\hat{j} \).

Since only the j-component of \( \vec{E} \) contributes to the flux, we have: \( \Phi = E_y \cdot A \), where \( E_y = 4 \times 10^3 \, N/C \).

Substituting the values, \( 6.0 = 4 \times 10^3 \times A \).

Solving for \( A \): \( A = \frac{6.0}{4 \times 10^3} = 1.5 \times 10^{-3} \, m^2 \).

Convert the area to \( cm^2 \): \( A = 1.5 \times 10^{-3} \, m^2 \times (10^4 \, cm^2/m^2) = 15 \, cm^2 \).

The area of the surface is 15 \( cm^2 \), which falls within the given range of 15,15, thus confirming the solution.

Answer 2

Step 1: Understand the orientation of the surface.
The rectangular surface is parallel to the x-z plane. This means that the normal vector to the surface is along the y-axis (either \( +\hat{j} \) or \( -\hat{j} \)). We can represent the area vector \( \vec{A} \) as \( \vec{A} = A \hat{j} \), where \( A \) is the area of the surface and \( \hat{j} \) is the unit vector in the y-direction.
Step 2: Use the formula for electric flux.
The electric flux \( \Phi \) through a surface is given by the dot product of the electric field \( \vec{E} \) and the area vector \( \vec{A} \): \[ \Phi = \vec{E} \cdot \vec{A} \]
Step 3: Substitute the given electric field and the area vector into the flux formula.
The electric field is \( \vec{E} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3 \, N/C \). The area vector is \( \vec{A} = A \hat{j} \). The flux is given as \( \Phi = 6.0 \, Nm^2C^{-1} \). \[ 6.0 = [(2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3] \cdot (A \hat{j}) \] The dot product of the unit vectors is \( \hat{i} \cdot \hat{j} = 0 \), \( \hat{j} \cdot \hat{j} = 1 \), and \( \hat{k} \cdot \hat{j} = 0 \). \[ 6.0 = (2 \times 10^3 \hat{i} \cdot A \hat{j}) + (4 \times 10^3 \hat{j} \cdot A \hat{j}) + (6 \times 10^3 \hat{k} \cdot A \hat{j}) \] \[ 6.0 = 0 + (4 \times 10^3 \times A \times 1) + 0 \] \[ 6.0 = 4 \times 10^3 A \]
Step 4: Solve for the area \( A \) in \( m^2 \).
\[ A = \frac{6.0}{4 \times 10^3} \, m^2 \] \[ A = 1.5 \times 10^{-3} \, m^2 \]
Step 5: Convert the area from \( m^2 \) to \( cm^2 \).
We know that \( 1 \, m = 100 \, cm \), so \( 1 \, m^2 = (100 \, cm)^2 = 10000 \, cm^2 = 10^4 \, cm^2 \). \[ A = 1.5 \times 10^{-3} \, m^2 \times \frac{10^4 \, cm^2}{1 \, m^2} \] \[ A = 1.5 \times 10^{-3 + 4} \, cm^2 \] \[ A = 1.5 \times 10^1 \, cm^2 \] \[ A = 15 \, cm^2 \] The area of the surface is \( 15 \, cm^2 \).