- #1
kiwileaf
- 4
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when deriving the electric field through using gauss's law, I do not completely understand why when
calculating for the charge enclosed the answer key says
"The surface charge distribution on is uniform. The area of the intersection of he non-conducting plane with the Gaussian cylinder is equal to the area of the end-caps, . Therefore the charge enclosed in the cylinder is q=σA"
I thought that the total charge would be the integral of sigma with respect to the volume of a cylinder (dv), which would make the total charge enclosed q=σV
However, even if it is q=σA I do not understand why we would not include the entire area of the cylinder because wouldn't the charge be over that entire Gaussian surface? why do we disregard the sides and only include the end caps for area?
calculating for the charge enclosed the answer key says
"The surface charge distribution on is uniform. The area of the intersection of he non-conducting plane with the Gaussian cylinder is equal to the area of the end-caps, . Therefore the charge enclosed in the cylinder is q=σA"
I thought that the total charge would be the integral of sigma with respect to the volume of a cylinder (dv), which would make the total charge enclosed q=σV
However, even if it is q=σA I do not understand why we would not include the entire area of the cylinder because wouldn't the charge be over that entire Gaussian surface? why do we disregard the sides and only include the end caps for area?