An electric field is given by \( (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C} \). The electric flux through a surface area \( 30\hat{i} \, \text{m}^2 \) lying in the YZ-plane (in SI units) is:
- 90
- 150
- 180
- 60
The Correct Option is C
Solution and Explanation
To determine the electric flux through a surface area in the presence of an electric field, we use the concept of the dot product between the electric field vector and the area vector. The formula for electric flux (\( \Phi \)) is:
\(\Phi = \vec{E} \cdot \vec{A}\)
Where:
- \(\vec{E}\) is the electric field vector
- \(\vec{A}\) is the area vector. The direction of \(\vec{A}\) is perpendicular to the surface.
In this problem, the electric field vector is:
\(\vec{E} = (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C}\)
The surface area is given as \(30 \hat{i} \, \text{m}^2\), and since it lies in the YZ-plane, the area vector is perpendicular to this plane. The direction of \(\hat{i}\) indicates that the area vector is:
\(\vec{A} = 30\hat{i} \, \text{m}^2\)
The dot product \( \vec{E} \cdot \vec{A} \) is calculated as follows:
\(\Phi = (6\hat{i} + 5\hat{j} + 3\hat{k}) \cdot (30\hat{i})\)
Using the properties of the dot product, where only the components in the same direction (\(\hat{i}\)) will contribute to the product, this simplifies to:
\(\Phi = 6 \cdot 30 = 180 \, \text{Nm}^2/\text{C}\)
Therefore, the electric flux through the surface is 180 Nm2/C.
This matches with the given correct answer option: 180.
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