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Let G = 4xa, +2zay+2ya,. Given an initial point P (2, 1, 1) and a final point Q(4, 3, 1), find f G-dL using the path: a) straight line: y = x-1, z = 1; b) parabola: 6y=x²+2, z = 1: With G as given, the line integral will be [G.dL = f* 4x s+ [₁³2zdy + f² 2ydz 4x dx + Clearly, we are going nowhere in z, so the last integral is zero. With z = 1, the first two evaluate as √ G. dL = 2x² | 2 + 2y| ²² = 28 The paths specified in parts a and b did not play a role, meaning that the integral between the specified points is path-independent.

Let G = 4xa, +2zay+2ya,. Given an initial point P (2, 1, 1) and a final point Q(4, 3, 1), find f G-dL using the path: a) straight line: y = x-1, z = 1; b) parabola: 6y=x²+2, z = 1: With G as given, the line integral will be [G.dL = f* 4x s+ [₁³2zdy + f² 2ydz 4x dx + Clearly, we are going nowhere in z, so the last integral is zero. With z = 1, the first two evaluate as √ G. dL = 2x² | 2 + 2y| ²² = 28 The paths specified in parts a and b did not play a role, meaning that the integral between the specified points is path-independent.

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Electromagnetic field)I want a detailed solution for this solution 11- Let G = 4xa, +2zay+2ya₂. Given an initial point P (2, 1, 1) and a final point Q(4, 3, 1), find f G-dL using the path: a) straight line: y = x-1, z = 1; b) parabola: 6y=x²+2, z = 1: With G as given, the line integral will be [G.dl. = [*Axdx + f2zdy + f zydz dL 4x Clearly, we are going nowhere in z, so the last integral is zero. With z = 1, the first two evaluate as √ G. dL = 2x² | 2 + 2y|²³ ² 12- The paths specified in parts a and b did not play a role, meaning that the integral between the specified points is path-independent. Three point charges, 0.4 µC each, are located at (0, 0, -1), (0, 0, 0), and (0, 0, 1), in free space. a) Find an expression for the absolute potential as a function of z along the line x = 0, y = 1: From a point located at position z along the given line, the distances to the three charges are R₁ = √√(z − 1)² + 1, R2 : 2 = √√₂² + 1, and R3 = √√(z + 1)² + 1. The total potential will be V(z) Using q = 4 x 10-7 C, this becomes V(z) = (3.6 × 10³) = = 28 9 1 1 1 + + 4л€0 R₁ R₂ R3 1 √(2−1)²+1 + 1 1 √2+1 √√2² +1 √(z+1)² + + V
Transcribed Image Text:Electromagnetic field)I want a detailed solution for this solution 11- Let G = 4xa, +2zay+2ya₂. Given an initial point P (2, 1, 1) and a final point Q(4, 3, 1), find f G-dL using the path: a) straight line: y = x-1, z = 1; b) parabola: 6y=x²+2, z = 1: With G as given, the line integral will be [G.dl. = [*Axdx + f2zdy + f zydz dL 4x Clearly, we are going nowhere in z, so the last integral is zero. With z = 1, the first two evaluate as √ G. dL = 2x² | 2 + 2y|²³ ² 12- The paths specified in parts a and b did not play a role, meaning that the integral between the specified points is path-independent. Three point charges, 0.4 µC each, are located at (0, 0, -1), (0, 0, 0), and (0, 0, 1), in free space. a) Find an expression for the absolute potential as a function of z along the line x = 0, y = 1: From a point located at position z along the given line, the distances to the three charges are R₁ = √√(z − 1)² + 1, R2 : 2 = √√₂² + 1, and R3 = √√(z + 1)² + 1. The total potential will be V(z) Using q = 4 x 10-7 C, this becomes V(z) = (3.6 × 10³) = = 28 9 1 1 1 + + 4л€0 R₁ R₂ R3 1 √(2−1)²+1 + 1 1 √2+1 √√2² +1 √(z+1)² + + V
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