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

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SUMMARY

The electric field at a distance of 5 cm from the center of two concentric insulating spherical shells can be determined using Gauss' Law. The inner shell, with a nonuniform charge distribution of 0.005 r C/m³, contributes to the enclosed charge through integration over its volume. The outer shell has a uniform charge distribution totaling 5 nC, which also contributes to the total enclosed charge. By applying Gauss' Law, the electric field can be calculated at the specified distance.

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zahraa
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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.

[tex]EA = \frac{q_{in}}{\epsilon_{0}}[/tex]
 

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