Questions: Consider an electric field right E = 2 i + 3 j. What is the magnitude of the flux of this field through a 4.0 m^2 square surface whose corners are located at (x, y, z) = (0,2,1), (2,2,1), (2,2,-1), (0,2,-1)?

Solution Steps

The given corners of the square surface are:

• \((0,2,1)\)
• \((2,2,1)\)
• \((2,2,-1)\)
• \((0,2,-1)\)

From these coordinates, we can see that the \(y\)-coordinate is constant at \(y = 2\). This indicates that the surface is parallel to the \(xz\)-plane and perpendicular to the \(y\)-axis.

Since the surface is perpendicular to the \(y\)-axis, the normal vector \(\hat{\mathbf{n}}\) to the surface is in the direction of the \(y\)-axis. The normal vector can be written as: \[ \hat{\mathbf{n}} = \hat{\mathbf{j}} \]

The electric field is given by: \[ \overrightarrow{\mathbf{E}} = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} \]

The dot product of \(\overrightarrow{\mathbf{E}}\) and \(\hat{\mathbf{n}}\) is: \[ \overrightarrow{\mathbf{E}} \cdot \hat{\mathbf{n}} = (2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}}) \cdot \hat{\mathbf{j}} = 3 \]

The flux \(\Phi\) through the surface is given by: \[ \Phi = \overrightarrow{\mathbf{E}} \cdot \hat{\mathbf{n}} \cdot A \] where \(A\) is the area of the surface. Given \(A = 4.0 \, \text{m}^2\), we have: \[ \Phi = 3 \cdot 4.0 = 12.0 \, \text{N} \cdot \text{m}^2/\text{C} \]

Final Answer