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Home/ Questions/Q 1043074
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junaid ansari
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junaid ansari
Asked: October 8, 20242024-10-08T08:48:22+05:30 2024-10-08T08:48:22+05:30In: Education

Evaluate Gauss law for D = 5r2 /4 i in spherical coordinates with r = 4m and θ = π/2

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      Quearn Quearnist
      2024-10-08T08:48:34+05:30Added an answer on October 8, 2024 at 8:48 am

      To evaluate Gauss’s Law using the provided electric flux density (D = frac{5r^2}{4} hat{i}) in spherical coordinates for (r = 4 ,m) and (theta = frac{pi}{2}), let’s first clarify the given parameters and the application of Gauss’s Law within this context.

      First, it’s important to note that the electric flux density (mathbf{D}) given as (D = frac{5r^2}{4} hat{i}) seems to be expressed in a form that mixes variables from Cartesian coordinates ((hat{i}) is a unit vector in the direction of the x-axis in Cartesian coordinates) with those of spherical coordinates ((r) is the radial distance in spherical coordinates). For a more consistent application related to Gauss’s Law in spherical coordinates, it’s essential to correctly interpret or transform this expression into spherical coordinates fully. However, assuming the intent was to express (mathbf{D}) in a purely spherical context and aiming to perform an evaluation as requested, one needs to correct the misunderstanding about the vector notations and how Gauss’s Law applies.

      —

      Given the context, let’s reframe the statement into a scenario Gauss’s Law could be applied to directly:

      If by (D = frac{5r^2}{4}) you meant the magnitude of the electric flux density in spherical coordinates without specifying a direction, and considering (

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    2. junaid ansari
      junaid ansari
      2024-10-08T08:48:57+05:30Added an answer on October 8, 2024 at 8:48 am

      c
      Explanation: ∫∫ ( 5r2
      /4) . (r2 sin θ dθ dφ), which is the integral to be evaluated.
      Put r = 4m and substitute θ = 0→ π/4 and φ = 0→ 2π, the integral evaluates to 588.9

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