For this problem on the topic. Off capacities and die electrics were given a solid conducting sphere of Radius, Capital R and Charge Cube. We want to calculate the electric field density inside the sphere so at a distance less than the radius and then outside the sphere at a distance greater than the radius. We also want to calculate the total electric field energy associated with the sphere. And we also want to explain why this this result can be interpreted as the amount of work that's required to assemble the charge on the sphere. Lastly, we want to show that the best INS is as given in the problems stated now. Firstly, we need to know that for a Southern conducting sphere of Radius R, the electric field is zero. Anywhere within the sphere on the electric field is Q over four pi excellent r squared anyway outside this year, so firstly, any distance are is less than the radius. So within this year we have the electric field density you, we know, is defined as a half times absolutely not. Times the electric field square the square. And since electric field is zero with this fear, electric field density is zero now for part B will do the same for any distance outside the radius of the sphere. Are is greater than capital are. So we have the electric field density again. You is a half absolutely not E squared, which is a half absolutely not. Into the charge. Q Divided by full pie, Absolutely not. R squared all squared. And so we get the clear energy density to be que squid over 32 pi squared. Absolutely not are square. And that's our electric field electric energy density anywhere outside the sphere. Now for part C, we want to calculate the total energy sources associated with electric fields of the total electric field energy, which is capital you. And this is simply the integral off the energy density over the volume. Now we know we can write Devi as four pi r squared the art So this integral becomes four pi, which is a constant times the integral everywhere outside this sphere. So from all to infinity, off r squared times you they are. And if we saw this integral, this becomes Q squared over eight pi absolutely not times the integral from our to infinity off the are by R squared and some things into go. We get the total energy the total electrical electrical energy associated with this fear to be a Q squared over eight pi. Excellent. Not times the radius off this year, Capital are, and that's the total energy associated with electric field. Now we want to show that this energy, calculated in part C he's actually the amount of work that's required to assemble the charge Q onto the spear. Now we can see that this energy, the total energy you associate electric field, these Q squared over a pie. Excellent, not times are, and so this energy is actually can be written as a half Q. Squared times four pi Absolutely not our, which looks familiar because this is the energy required to assemble all the charge into hysterical distribution. Note that we have to be aware of double counting, so this factor off half in front off the potential. The familiar potential energy formula for a charge Q and A distance are from any other charge Magnitude Cube. And so this is essentially the work done in assembling the charge onto this fear. Now, finally, we want to calculate the capacities. We went to find that the capacitance on this fear is has given as the problems stated. So we know that the energy you associated with electric field is Q squared over twice the capacitance to see. And from above we said You is equal to Q squared over eight pi. Excellent, not times are so Therefore we can see by equating these two expressions and rearranging, we get the capacitance C to some TV for pie. Absolutely not. Terms are as the quiet.