A thin spherical shell with radius R_1 = 3.00 cm is concentric with a larger thin spherical shell with radius R_2 = 5.00 cm. Both shells are made of insulating material. The smaller shell has charge q_1 = +6.00 nC distributed uniformly over its surface, and the larger shell has charge q_2 = -9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) r=0; (ii) r=4.00 cm; (iii) r=6.00 cm?

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Hey, everyone. Today, we're dealing with a problem about electric potential and electric potential energy. So we're being told that we have to hollow plastic globes with radius of 4.5 cm and 8.4 cm respectively. And these two hollow globes are Cohen centric with each other. Now, the globe of Radius 4.5 cm has a charge of negative 12.5 Nano columns distributed evenly on its surface. Similarly, the other globe has a charge of 8.25 Nano columns evenly distributed over its surface. If the electric potential is zero at an infinitely large distance, if electric potential zero at an infinitely large distance, we're being asked to determine the resultant electric potential of the two globes at one where the radial distance from the center of the shell is zero centimeters Where it is six cm and where it is 8. cm. So let's write down everything we know so far, we know that the, and let's use one to denote anything related to the inner circle or inner sphere and the number two to denote anything related to the outer sphere. So we know that Q one the inner sphere has a charge of negative 12. Nano columns, nano columns and the radius of the shell will use capital R Is 4.5 cm. However, we want to convert these to our standard S I units. So we can recall a couple of conversion factors. Oops, excuse me, we can recall a couple conversion factors. One that one nano column is equal to 10 to the negative ninth columns. And the second is that one centimeter Is equal to 10 to the night of 2nd m. So we can rewrite these as negative 12.5 times to the -9 columns. And the radius will be 0.045 m. Similarly, we can use to didn't know the outer sphere. Thank you. So this would be, well, it is uh positive 8.25 nano columns. So we can convert that that will be positive 8.25 times 10 to the negative nine columns. And excuse me, and the radius radius of two Is 8.4 cm or 0. m. So with all this in mind, let's go ahead and realize that potential due to a shell can also be expressed as fee. The potential is equal to K which is cool, it's constant multiplied by the charge divided by our larger the radius of the sphere. If are the radial distance from the center of the sphere that we're looking for or that were uh including into our calculations is less than the radio, our radius from the center of the sphere. Or we can say that the is equal to K Q over a lower case R when R or when uh sorry R or lowercase artists greater than capital R or when the radio distance from the center of the sphere is greater than the radius of the sphere itself. So with this in mind, let's go ahead and take a look at scenario one, oops oops that's a little out of frame, Look at Scenario one. So we know that the potential at any point is a scalar some of the potential from each shell. Now, the positive charges here create a positive potential around them and the negatives create a negative potential around them. So we need the sign of a charge when substituting it into the expression. And what does this mean while the total sum of all potentials should be V one plus 32, because again, the potential at any point is the scalar some of the potential from each shell. So here in example, one or in scenario one lower case R, the radial distance from the center of the sphere is zero centimeters. So R is less than the radio distance are the radius of the spheres. So with that in mind, we can go ahead and use this first case. So we know that If we substitute in our here, we need to calculate both for R one and R two. So we can say uh the is equal to que Q one over, excuse me, KQ one over R one, R one Plus KQ two over R two, which can be further simplified to K Q one over R one Plus Q two over R two. So with this in mind, we can go ahead and just plug in our values putting in our values gives us nine times 10 to the ninth Newton meters squared over Coolum square. This is cool ems constant, constant K and then we'll have um excuse me, negative 12.5 times 10 to the -9 columns over Our one is 4 or zero 045 m. We have Q2 which is 8.25 times 10 to the 99th columns divided by 0.084 m which gives us a value of negative 1.61 times 10 to the negative three volts Or negative 1.61 or oops, excuse me, not negative or 1. kilovolts there at that little neater 1.61 kilovolts. So we'll highlight that. So that is the resultant electric potential of the two globes. When the radial distance R at the distance of zero at a radial distance from the center of the sphere, zero or at the center of both spheres, I'm gonna scroll down. So we have a little more space Case two is very similar, right Case two is similar where the radial distance is now greater than the first, um not greater than the first fear, but smaller than the second fear or second sphere, excuse me. So we can go ahead and factor that in, we can use similar logic processes as we did. For the first one, There's six cm between the globes. So we can say that we have V one or V is equal to V one plus V two where R is equal to 0.0 six m and are will be art too. So substituting these into our equation, this gives us K Q one over R because now for Q one R is greater than the first fear which is 4.5 centimeters. So we simply just use lower case R. This will be Q two over R two. So this will give us nine times 10 to the 9th meters squared over Coolum square. It's native 12.5 times 10 to the NATO ninth Coolum over 0.6 m plus 8.25 times 10 to the ninth. I messed up my calculations above earlier. It's not negative nine is positive or whoops. No, I did right the first time, I bet I got confused with cool number. No, this is negative 8.25 times 10 to the negative ninth columns because this is 8.25 nano columns over 0.084 m which gives us an answer of negative 990 killer or negative 990 volts. My apologies, this is negative 1.61 kilovolts. I forgot to add that earlier or negative 0.99 kilovolts. So we can use a similar process yet again for the final scenario. However, this time, this time, the value of our lower case R is the same as our two. So we'll just factor that in and we'll continue salami. So we have three, V is equal to V one Plus V two and R is equal to R two here. So we use the exact same as we did up here. We'll use the same principle as we used up here that V is equal to K into Q one over R lowercase R plus Q two over R two since they're equal. So solving for this, we get nine times 10 to the ninth Newton two m squared over cool um squared. Well, it's applied by negative 12.5 times 10 to the native ninth columns over 0. meters plus 8.25 times to the NATO ninth columns over 0. m which gives us an answer of negative 445 or sorry, eight and 455 volts or negative 0.455 kilovolts. So this means the resultant electric potentials from the two globes. When lower case R is equal to zero centimeters, six centimeters and 8.4 centimeters is negative 1.61 kilovolts, negative 0.99 kilovolts and 0.0 or 0. kilovolts or answer choice C respectively. I hope this helps and I look forward to seeing you all in the next one.

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Key Concepts

Electric Potential

Gauss's Law

Superposition Principle