The inner sphere has a net charge of +6 μC and the shell has a net charge of +6 μC.

Name: Problem 60, Chapter 16: Physics
Uploaded: 2021-07-16T16:14:39-08:00
Duration: 6 min 41 s
Channel: Jose Carlos
Description: The inner sphere has a net charge of +6 μC and the shell has a net charge of +6 μC.

The inner sphere has a net charge of +6 μC and the shell has...

Key Concepts

Gauss's Law

Gauss's law relates the electric flux through a closed surface to the charge enclosed by that surface. This concept is crucial for calculating the electric fields produced by symmetric charge distributions, such as those found in spherical systems, by reducing complex integrals to simpler forms based on symmetry.

Spherical Symmetry

Spherical symmetry simplifies the analysis of electric fields and potentials since the symmetry ensures that the field has the same magnitude at every point equidistant from the center. This principle is fundamental when applying Gauss's law to determine the behavior of fields arising from spherically distributed charges.

Charge Distribution in Conductors

In conductors, charges redistribute themselves in response to electric fields until equilibrium is reached, resulting in no electric field within the bulk of the material. This concept is especially important for understanding how charges reside on the surfaces of objects like spheres and shells and how the distribution affects the resulting electric fields.

Electrostatic Induction

Electrostatic induction describes the process by which a charged body causes a redistribution of charges in a nearby conductor, even without direct contact. This principle is critical in systems where multiple objects interact electrically, influencing the local charge distribution and consequent electric fields.

Superposition Principle

The superposition principle asserts that the resultant electric field created by multiple charge distributions is equal to the vector sum of the fields produced by each individual charge. This concept is essential for analyzing systems with more than one charged entity, as it allows for the independent treatment of each charge's contribution before summing them up.

Transcript

00:01 In this problem, we have a sphere surrounded by a shell, and in this part of the problem, each of them carries positive 6 microcouple.

00:11 Part a requires us to draw the field lines, and to do that we need to follow two principles.

00:18 One of them is that field lines go from positive charges to negative charges.

00:25 So that means away from positive charges.

00:29 The second idea is that there cannot be field lines inside conducting materials.

00:39 That means that for the sphere our field lines have to start at the surface of the sphere and they have to point away from it.

00:48 And they also have to stop once they get to the shell because the shell is also a conducting material.

00:55 So we have them like that.

00:59 But it's not like these field lines disappear here.

01:02 They continue once they leave the shell.

01:08 So we have them like this.

01:11 But let's not forget that the shell also has some charge.

01:15 So that means that the shell is going to create its own field lines as well.

01:20 And since it has positive six columns as well, we need to draw four field lines, for additional field lines.

01:33 There we go.

01:34 Here we have our diagram.

01:37 Now for part b it is useful to look at a clean diagram once again.

01:43 And we also need to recall gosses law.

01:47 Gus's law states that the integral of the electric field times with respect to area is equal to the charge enclosed over permittivity of free space.

02:05 Now, if we don't have an electric field within the shell, then that means that there must be some charge canceling out the inner spheres charge.

02:19 So the inner spheres charge has positive microculems, 6 microculems.

02:33 That means that the inner portion of the shell must have negative 6 microculems so that they cancel out this inner positive 6.

02:45 Now in the outer portion, you must think, well, if we have positive 6 and we put positive 6, then they add up to 0...

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