Fatima Hasan
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The area that we're concerned will be the surface area of the ends of Gaussian surface which is equals to π / R^2TSny said:Which parts of the surface of the cylinder have nonzero flux?
Of course you mean π⋅R^2.Fatima Hasan said:The area that we're concerned will be the surface area of the ends of Gaussian surface which is equals to π / R^2
This is correct.Φ = E A = Q enclosed / ε
Q enclosed = 100 * π * (0.1)^2 * 8.85 = 27.8 pC
##A = 2 \pi r h## , we use this formula to find the net flux through a cylinder , right ?TSny said:Of course you mean π⋅R^2.
This is correct.
Not in this problem. The area ##A = 2 \pi r h## is the "lateral" area of the curved surface of the cylinder, as shown below in blueFatima Hasan said:##A = 2 \pi r h## , we use this formula to find the net flux through a cylinder , right ?
NoTSny said:Not in this problem. The area ##A = 2 \pi r h## is the "lateral" area of the curved surface of the cylinder, as shown below in blue
View attachment 221399
Is there any flux through the blue surface in the problem you are working on?