How to Calculate Electrostatic Potential in a Uniformly Charged Spherical Layer?

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To calculate the electrostatic potential inside a uniformly charged spherical layer, one can utilize the surface charge density and apply a surface integral over the charge distribution. The potential can be expressed as a volume integral, but due to the spherical symmetry, it simplifies to a surface integral. Gauss's law can be employed for a quick solution, but if it's not permitted, focusing on the polar axis (phi=0) is sufficient since the potential remains uniform regardless of theta or phi. Understanding the relationship between surface area and volume in spherical coordinates is crucial for setting up the integral correctly. The electrostatic potential inside the sphere is uniform at all points.
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Hello

i have a problem with spherical coordinates however i understand partial differentials.

Basically a thin spherical layer of radius R is uniformly charged with surface charge density. Show by Calculating the electrostatic potential (x,y,z) produced by the layer that electrostatic potential is unifrom anywhere inside the sphere.Use spherical coordiantes.It also says a differential area da=R^2sin()d()d(/) can be obtained from the volume dv=R^2 sin()Rd()d(/)=dAdR

()=theta
(/)=phi

I know how to calculate the moment of inertia using spherical coordinates however i don't know how to start off the question. Am i suppose to calculate the are or the volume and what equation do i put into the triple integral.Very confusing?:bugeye:
 
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The equation for the potential is a volume integral over the charge distribution. In this case, you only have surface charge, so this turns into a surface integral of the surface charge density divide dy |r-r'|, where r is the position you are calculating the potential at and r' is the position of the charge (you integrate over r'). In this case, because of the spherical symmetry, you could use Gauss's law to solve this problem in one line. If you don't know Gauss's law, or aren't allowed to use it, you can still simplify the problem considerably by noting that you only need to consider r along the polar axis (phi=0), since the potential shouldn't depend on theta or phi.
 

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