
A cylinder with circular cross-section is the appropriate Gaussian surface to use when analyzing all of the following charge distributions except
A. A long, straight charged wire
B. A point charges.
C. The inside of a uniformly charged, insulated rod
D. The inside of a charged, insulated rod whose charge density varies with radius
E. A large, uniformly charged plane
Answer
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Hint: The flux of the electric field $E$ through any closed surface, also known as a Gaussian surface, is equal to the net charge enclosed (${q_{enc}}$) divided by the permittivity of free space (0) according to Gauss's law: Gauss's law determines the electric flux through any closed surface surrounding a point charge q.
Complete answer:
As discussed above the electric field is radially outwards for a point charge. Here in the given options for options A, C, D and E the gaussian surface leads to cylindrical form as by drawing the direction of the electrical field in each of the options. But in option B the gaussian surface is a sphere as a charge is a point charge so this leads to a sphere.
Hence, option B is the correct answer.
Note:Select a gaussian surface that passes through the point at which the electric field is desired. The ideal surface is one where the electric field is constant in magnitude and makes the same angle with the surface at all times, making the flux integral easy to calculate.
Complete answer:
As discussed above the electric field is radially outwards for a point charge. Here in the given options for options A, C, D and E the gaussian surface leads to cylindrical form as by drawing the direction of the electrical field in each of the options. But in option B the gaussian surface is a sphere as a charge is a point charge so this leads to a sphere.
Hence, option B is the correct answer.
Note:Select a gaussian surface that passes through the point at which the electric field is desired. The ideal surface is one where the electric field is constant in magnitude and makes the same angle with the surface at all times, making the flux integral easy to calculate.
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