What will the electric field be at point at a distance r=5cm

AI Thread Summary
To find the electric field at a distance of 5 cm from the center of two concentric insulating spherical shells, one must calculate the enclosed charge using Gauss' Law. The inner shell has a nonuniform charge distribution of 0.005 r C/m^3, which requires integration over its volume to determine the total enclosed charge. The outer shell has a uniform charge distribution with a total charge of 5 nC, and only the portion within the Gaussian surface at 5 cm contributes to the total charge. By summing the charges from both shells and applying Gauss' Law, the electric field can be derived. This approach effectively combines the contributions from both shells to determine the electric field at the specified distance.
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Homework Statement


A small insulating spherical shell with inner radius a=1cm and outer radius b=2cm is concentric with a larger insulating soherical shell with inner radius c=4cm and outer radius d=6cm . the inner shell has a nonuniform charge distribution =.005 r C/m^3 , and the outer shell has a uniform charge distribution and a total charge of 5 nC .

Homework Equations



my question is what will the electric field be at boint at a distance r=5cm from the common center of both spheres

The Attempt at a Solution



q/(4*pi*E0*(b-a)^2) ??
 
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Hey zahraa,

Just determine the volume enclosed by the portion of the outer shell that is contained within your Gaussian surface. Use that to get the portion of the charge enclosed by that part of the shell.

Then for the nonuniform part of the enclosed charge form an integral over the volume to get the portion of the charge enclosed by the inner shell's volume.

Add those enclosed charge values and use Gauss' Law for the Gaussian shell with radius 5cm.

EA = \frac{q_{in}}{\epsilon_{0}}
 
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