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 (a) the volume charge density for the sphere?

Video transcript

Hey, everyone. So this problem is dealing with electric fields. Let's see what it's asking us. We have a solid ball made of an insulating material that has a negative charge of Q spread uniformly over its volume. The radius of the ball is given us six centimeters and the magnitude of the electric field measured 10 centimeters from the center of the ball is 245 newtons per Coolum. We're asked to find the volume charge density given by row for the insulating ball are multiple choice answers here in units of Coolum per meter, cubed are a negative 3.01 times 10 to the negative seven B negative 2.4 times 10 to the negative eight C negative 5.41 times 10 to the negative nine or D negative 2.72 times 10 to the negative 10. OK. So the first thing we can do is recall that our volume charge density equation is given by row equals Q divided by V. We have the um radius and can use that to find our volume. But we don't have that value of Q. So we can recall that the electric field, a point charge is equal to E equals K absolute value of Q divided by R squared. We can use this equation because the um electric field magnitude was measured outside of the sphere. So the ball only has a radius of 10, of six centimeters, so 10 centimeters is outside of that ball. And so we can treat that entire ball made, uh treat that entire ball as a point charge. So rearranging this equation, we have Q equals er squared divided by K and we can plug in the values from the problem tool for that point charge. So our electric field magnitude newtons per Coolum multiplied by the radius. So this is the radius where that electric field was measured. So that's going to be 10 centimeters, keeping everything in standard units. I'm going to write that as 0.1 m, the radius squared all divided by K which is our Coolum constant. We can recall is 8.99 times 10 to the nine new meters squared per Coolum squared. And so when we plug all of that into our calculator, we get a charge of 2. times 10 to the negative 10 colons. Now we were told that it has a negative charge. So Q is going to be negative 2.72 times 10 to the negative 10 colons. OK. Looking at our volume charge density equation. Now we have Q we can find our volume. Recalling that the volume for a sphere for a sphere is given by four cubes pi R or sorry four thirds pi R cubed where the radius of the ball was given as six centimeters. So standard units will keep that as 0.06 m. So four thirds multiplied by pi multiplied by 0.06 m cubed equals 9. times 10 to the minus 4 m qed. And so then our last step is to solve for that volume charge density. And we have Q which was negative 2. times 10 to the negative 10 columns divided by 9.05 times 10 to the negative 4 m cubed. And that comes out to 3.01 times 10 to the negative seven columns per meter cubed. And that is our final answer. When we look at our multiple choice answers that aligns with answer choice A so that's all we have for this one. We'll see you in the next video.