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Question 2: Gauss's Law: Calculus based For this problem assume constants are all in SI units that are not shown (a) An electric field given by E = (3x²). Calculate the flux, PE = $E· dà = f S (E cos(0))dx dy, through surface S shown to the right. The surface exists in the x-y plane, has one corner on the origin (0,0,0), and has side length a = 10cm. (b) Repeat the preceding problem, with E = (2x)i + (4x³) k a a 425

Question 2: Gauss's Law: Calculus based For this problem assume constants are all in SI units that are not shown (a) An electric field given by E = (3x²). Calculate the flux, PE = $E· dà = f S (E cos(0))dx dy, through surface S shown to the right. The surface exists in the x-y plane, has one corner on the origin (0,0,0), and has side length a = 10cm. (b) Repeat the preceding problem, with E = (2x)i + (4x³) k a a 425

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**Question 2: Gauss’s Law: Calculus Based** For this problem, assume constants are all in SI units that are not shown. (a) An electric field given by \(\vec{E} = (3x^2) \hat{k}\). Calculate the flux, \(\Phi_E = \int \vec{E} \cdot d\vec{A} = \int \int (E \cdot \cos(\theta)) dx \, dy\), through surface \(S\) shown to the right. The surface exists in the x-y plane, has one corner on the origin (0,0,0), and has side length \(a = 10 \text{cm}\). (b) Repeat the preceding problem, with \(\vec{E} = (2x) \hat{i} + (4x^3) \hat{k}\). **Diagram Explanation:** The diagram depicts a coordinate system with axes labeled \(x\), \(y\), and \(z\). A square surface \(S\) is oriented in the x-y plane with corners aligned along the axes. Each side of the square is denoted by a length \(a\). The surface position on the x-y plane shows it has one corner at the origin (0,0,0). Vectors indicate components along the axes, with the surface normal visibly pointing upwards, suggesting the \(z\)-component is of most interest in the flux calculation.
Transcribed Image Text:**Question 2: Gauss’s Law: Calculus Based** For this problem, assume constants are all in SI units that are not shown. (a) An electric field given by \(\vec{E} = (3x^2) \hat{k}\). Calculate the flux, \(\Phi_E = \int \vec{E} \cdot d\vec{A} = \int \int (E \cdot \cos(\theta)) dx \, dy\), through surface \(S\) shown to the right. The surface exists in the x-y plane, has one corner on the origin (0,0,0), and has side length \(a = 10 \text{cm}\). (b) Repeat the preceding problem, with \(\vec{E} = (2x) \hat{i} + (4x^3) \hat{k}\). **Diagram Explanation:** The diagram depicts a coordinate system with axes labeled \(x\), \(y\), and \(z\). A square surface \(S\) is oriented in the x-y plane with corners aligned along the axes. Each side of the square is denoted by a length \(a\). The surface position on the x-y plane shows it has one corner at the origin (0,0,0). Vectors indicate components along the axes, with the surface normal visibly pointing upwards, suggesting the \(z\)-component is of most interest in the flux calculation.
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