Question

Given a uniform electric field \[\vec{E} = 2 \times {10}^3 \ \hat{i}\]  N/C, find the flux of this field through a square of side 20 cm, whose plane is parallel to the y−z plane. What would be the flux through the same square, if the plane makes an angle of 30° with the x−axis ?

Answer 1

When the plane is parallel to the y-z plane:

\[\phi = \vec{E .} A^\rightharpoonup \]

\[Here: \]

\[ \vec{E} = 2 \times {10}^3 \ \hat{i} N/C\]

\[ A^\rightharpoonup = \left( 20 cm \right)^2 \hat{i} = 4 \times {10}^{- 2}  \ \hat{i} m^2 \]

\[ \therefore \phi = \left( 2 \times {10}^3 \ \hat{i} \right) . \left( 4 \times {10}^{- 2} \ \hat{i} \right)\]

\[ \Rightarrow \phi = 80 \ \text { Weber }\]

When the plane makes a 30° angle with the x-axis, the area vector makes a 60° angle with the x-axis.

\[\phi = \vec{E .} A^\rightharpoonup \]

\[ \Rightarrow \phi = EA \cos\theta\]

\[ \Rightarrow \phi = \left( 2 \times {10}^3 \right)\left( 4 \times {10}^{- 2} \right)\cos60°\]

\[ \Rightarrow \phi = \frac{80}{2}\]

\[ \Rightarrow \phi = 40 \text { Weber}\]