Given a uniform electric filed \[\vec{E} = 4 \times {10}^3 \ \hat{i} N/C\]. Find the flux of this field through a square of 5 cm on a side whose plane is parallel to the Y-Z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis?

#### Solution

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

\[\text { Electric flux }, \phi = \vec{E .} A^\rightharpoonup \]

\[\text { Here }: \]

\[ \vec{E} = 4 \times {10}^3 \ \hat{i} N/C\]

\[ A^\rightharpoonup = \left( 5 cm \right)^2 \ \hat{i} = 0 . 25 \times {10}^{- 2} \ \hat{i} m^2 \]

\[ \therefore \phi = \left( 4 \times {10}^3 \ \hat{i} \right) . \left( 25 \times {10}^{- 4} \ \hat{i} \right)\]

\[ \Rightarrow \phi = 10 \text { Weber }\]

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

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

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

\[ \Rightarrow \phi = \left( 4 \times {10}^3 \right)\left( 25 \times {10}^{- 4} \right)\cos60°\]

\[ \Rightarrow \phi = \frac{10}{2}\]

\[ \Rightarrow \phi = 5 \text { Weber }\]