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# Suppose the conducting spherical shell of Figure 15.29 carries a charge of 3.00 $\mathrm{nC}$ and that a charge of $-2.00 \mathrm{nC}$ is at the center of the sphere. If $a=2.00 \mathrm{m}$ and $b=2.40 \mathrm{m}$ , find the electric field at (a) $r=1.50 \mathrm{m},(\mathrm{b}) r=2.20 \mathrm{m},$ and $(\mathrm{c}) r=2.50 \mathrm{m}$ (d) What is the charge distribution on the sphere?

## a. $-7.99 \mathrm{N} / \mathrm{C}$b. $E=0$c. 1.44 $\mathrm{N} / \mathrm{C}$d. 1.0 $\mathrm{nC}$

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##### Christina K.

Rutgers, The State University of New Jersey

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg

### Video Transcript

in this problem. On the topic of electric forces and fields, we are given a conducting spherical shell in the figure, which carries a charge of three Nana columns with a charge of minus two Nana columns at its center. We want to find the electric field at 1.5 m 2.2 m and 2.5 m from the center. We also want to know the charge distribution on the sphere. So if we draw our guardians, fear or Gaussian surface as follows and note that the point charge minus two nano columns is positioned at the center of the spherical shell. We have complete spherical symmetry in this situation, so we can expect the distribution of charge and the shell, as well as electric fields both inside and outside of the shell to be physically symmetric. So let's first choose this vehicle calcium surface centered on the center of the conducting shell, where the radius R is equal to one 0.5 m less than a so we can use gases law and first find the electric flux. Five e is equal to the electric field magnitude times the spherical surface area A which we can write as e times for hi R squared and we know this from God's law to be the charge at the center of the enclosed charge. Q. Inside times, absolutely not, and this implies that the electric field is equal to he. Enclosed charged, Enclosed charge Q. Inside divided by four pi Absolutely not r squared. And we know one of the four, perhaps or not, is the electric constant. K E times Q. Inside, which is the child at the center, divided by our square so we can apply this formula and find the electric field at 1.5 m. So Kay, that's one or 8.99 times 10 to the power nine Newton meter squared. Oh, cool, um, squared times the charge, which is only due to the minus two Nano column charge, which is minus two times 10 to the minus nine columns divided by R squared, which is 1.5 m squared. So calculating we get the electric field to B minus seven 0.99 Newton's cooler, and the negative sign here tells us that this electric field is readily inward. That's 7.99 Newtons. Cool. Um, readily inward. So that's the electric field strength at a distance of 1.5 m from the to Nana column charge. Next for part B, we want to find the electric field strength at 2.2 m from the center. So R is equal to two 0.2 m. Means that our lives between A and B Now, since we are located within the conducting material making up the shell, we know that under the conditions of electrostatic equilibrium, the field is equal to zero at all points inside a conducting material. The electric field strength inside is equal to zero. Now, lastly, we want to find the electric field strength when R is equal to two 0.5 m and this lies outside B so we can use gas is low again with total spherical symmetry to lead us to as above e is equal to okay times the enclosed charge over r squared but now enclosed charge you inside consists of the charge at the center and the charge on the shelves of the total charge of the shower. You shall plus the minus two Nano column chart at the center. The total charge on the shell is positive. Three Nana columns and on the charge in the center is minus two nano columns, so the total net charge is positive. One Nana columns enclosed at R is equal to 2.5 m. So therefore, the electric field strength E if we substitute our values and articulation above is 8.99 times 10 to the nine Newton meter squared. McCullum squared times the charge, which is one Nana column, or one times 10 to the minus nine columns divided by 2.5 meters squared. And so we get electric field strength outside of B two b positive 1.44 Newton's curriculum and the positive science tells us that this field is pointing readily outward, so that's the electric field. Strength at R is equal to 2.5 m. Lastly, in party, we want to know the charge distribution on the sphere. Now under the conditions of electrostatic equilibrium, all excess charge and a conductor resides entirely on its surface. So the some of the charge and the inner surface of the shell, and that on the outer surface of the shell, Hugh Shell is positive. Three Nana columns to see how much of this is on the inner surface? We have to consider our calcium surface to have a radius R as infinite as many larger than a so all points in the gaseous surface lie within the conducting material meaning that is equal to zero at all points. Total flux to the surface five is equal to zero. So Carsons Castle Law state that mm totally enclosed charge you inside is equal to Q on the inner surface was the charge at the center, and this was equal to zero, meaning that the charge contained on the inner surface is equal to minus the charge at the center. And we know this to be minus of minus two Nana columns. So this is simply positive to Nana columns. So we charge on the outer surface. Must be. We'll call it Q. Outer is equal to the charge on the shell. Miners did charge on the inner surface, and that's three nano columns. Mine is positive. Two Nano columns, which gives us one nano Coolum

University of Kwazulu-Natal
##### Christina K.

Rutgers, The State University of New Jersey

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg