Charge q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00 cm from the center of the sphere, the electric field due to the charge distribution has magnitude E = 940 N/C. What are (b) the electric field at a distance of 2.00 cm from the sphere's center?

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Hey, everyone. So this problem is dealing with electric fields. Let's see what it's asking us. We have a uniform glass globe that acquires a positive charge Q that is distributed evenly across the globe's volume. The globe has a radius of 15 centimeters and the strength of the electric field just outside the globe is the 350 newtons per Coolum. Calculate the electric field at 10 centimeters from the globe center. Our multiple choice answers here in Newton's per Coolum are a 74.5 B 233. C 700.2 or D 744.6. OK. So the first thing that we can do is find the electric field outside of the globe because outside a spherically symmetrical charge, we can treat it as a point charge. So we can recall that that equation is E equals K multiplied by the absolute value of Q divided by R squared. And so given that we have the um electric field, the radius K is of course, Kums constant, we can solve for the absolute value of the charge right outside the sphere. And so when we rearrange the that equation, we get er squared divided by K. And so plugging in those values newtons per Coolum multiplied by the radius. It was given as 15 centimeters. We want to keep that in standard units. So we'll rewrite that as 0.15 m squared. And then we can recall that Kums constant is 8.99 times 10 to the nine newton meters squared per squared. And so when we plug that in to our equation, we get a charge of 8.76 times 10 to the negative cool ups. And now they're asking for the electric field 10 centimeters from the globe center. So while we found the charge outside, we can use that to help us find the electric field inside by using are charge density which will not change whether it's inside or outside of the. Um see you. So for Gaussian surfaces, we can recall that the enclosed charge divided by epsilon knot is equal to energy multiplied by area. And so rearranging that equation for sorry, not energy, the ele sorry, the electric field electric field multiplied by area. So rearranging that equation for the electric field, we get Q enclosed divided by alpha epsilon knot. So we have queue outside of the sphere outside of the globe, we need to find this queue enclosed. And we can again, we can do that using the um charge density. So the charge density is going to be equal to the charge or Q divided by the volume. So when we have a charge of 8.76 times 10 to the negative columns, that is at a radius of centimeters, we can recall the volume of a sphere is given by four C four Q four thirds pi R cubed. And again, that radius is 0.15 m. And so that gives us a um charge density of 6. times 10 to the negative eight columns per meter cubed. OK. So our next step then is to find Q enclosed at the point that the problem is asking about, which is when the radius is 0.1 m, 10 centimeters is 0.1 m um versus this previous charge which was at uh centimeters. So we can write that as Q and closed is equal to row V where we found row. So that's 6.2 times 10 to the negative eight columns per meter cubed. And that's multiplied by the volume. So uh at the point in question, which is four thirds pi and a radius of 0.1 m squared or sorry cubed. And when we plug that in, we get a queue enclosed of 2.5 sorry five, 97 times to the negative 10 cool ups. OK? And so now we have our queue enclosed. Our last step is going to be just plugging in these values and solving for our electric field. So we have Q and close 2.597 times 10 to the negative 10 colons divided by our area. So our um surface area is given by pi R squared, four pi R squared. So four pi again, that um radius is 0.1 m squared and then multiplied by epsilon knot, which is a constant. You can recall 8.854 times 10 to the negative 12 Pullum squared per Newton meters squared. And we plug all of that in and we get an electric field of 233.4. And that's units Newton per Coolum. And when we go back up to our multiple choice answers that aligns with answer choice B so B is the correct answer for this problem. So that's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Electric Field

Gauss's Law

Uniform Charge Distribution