Electric field in the Spherical Cavity

[Physicslearner500039]

Homework Statement

A nonconducting solid sphere has a uniform volume charge density p. Let r be the vector from the center of the sphere to a general point P within the sphere. (a) Show that the electric field at P is given by If = p7I3so. Fig. 23-56 (Note that the result is independent of the Problem 73. radius of the sphere.) (b) A spherical cavity is hollowed out of the sphere, as shown in Fig. 23-56. Using superposition concepts, show that the electric field at all points within the cavity is uniform and
equal to If = pa/3so, where a is the position vector from the center of the sphere to the center of the cavity

Relevant Equations

NA

a. For the question a the solution is
If the uniform charge density is ρ then the charge of the sphere up to radius r is
q = ρ * (4/3)*π * r3;
Hence the electric field is
E = (ρ *4π*r^3)/(3*εο*r^2); E = (ρ*r)/(3εο);
b. I don't understand what is superposition? How to proceed? Please advise.

 

[acchyut_jolly]

By superposition, what is meant here is that the cavity given to you can be considered as a sphere of charge density negative of that of the larger sphere. So basically you have to consider a negatively charged sphere superposing with the larger positively charged one only in the region where you are given the cavity. When you solve by adding the electric fields of both the spheres inside the cavity, you will get your required result.

 

[Physicslearner500039]

I did not understand completely. When you made a cavity you basically removed the charge from that portion.

 

[hutchphd]

...Or you could think about it as adding a sphere of opposite charge density to that region. Because of superposition (linearity) the solution to the problem is the appropriate sum of the easier individual solutions.

 

[archaic]

Here's what they are trying to tell you:
$$\mathrm E_{total}=\mathrm E_{sphere}-\mathrm E_{cavity}=\mathrm E_{sphere}+\left(-\rho\times\text{the rest of the terms in }\mathrm E_{cavity}\right)$$
You interpret the field of the cavity as shown in the last addition as a sphere having the same but opposite charge density.
The superposition theorem is that if you have two "things" from which results two electric fields, then the electric field at each point is the sum of the two electric fields. You can extend this to $n$ discrete things, or a continuous distribution as is your sphere :)

 

[etotheipi]

...Or you could think about it as adding a sphere of opposite charge density to that region. Because of superposition (linearity) the solution to the problem is the appropriate sum of the easier individual solutions.

I might add just for clarity that the final field strength is that due to the initial full sphere of charge added to that due to a negated sphere of charge at the location of the cavity, and not the field strength obtained if we filled the cavity with material of negated charge density - as in, it's an algebraic trick.

This is what yourself and @acchyut_jolly were saying, but just want to spell it out just in case it helps anyone .

 

[Physicslearner500039]

Ok, so this is my understanding, please correct me if i am wrong, how the charges are replaced will be something like this.

The calculations are something like this.

Are the above understanding and calculations correct? Please advise.

 

[hutchphd]

The diagrams are difficult for me to understand in detail. The equations are correct (so long as we agree that a,r,and E are vectors) and so I think you get the idea. A surprising result (to me at least) but looks correct. The superposition idea (and the similar method of images) are very very useful, so understand them well.