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प्रश्न
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 ?
उत्तर
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.
\[\phi = \vec{E .} A^\rightharpoonup \]
\[ \Rightarrow \phi = EA \cos\theta\]
\[ \Rightarrow \phi = \left( 2 \times {10}^3 \right)\left( 4 \times {10}^{- 2} \right)\cos60°\]
\[ \Rightarrow \phi = \frac{80}{2}\]
\[ \Rightarrow \phi = 40 \text { Weber}\]
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