FIGURE P25.67 shows two uniformly charged spheres. What is the...

Question

FIGURE P25.67 shows two uniformly charged spheres. What is the potential difference between points 1 and 2? Which point is at the higher potential? Hint: The potential at any point is the superposition of the potentials due to all charges.

Accepted Answer

Hey, everyone in this problem, we're provided a diagram that illustrates an arrangement of two uniformly charged spherical balls. We're asked to determine the potential difference VM minus VN between positions M and N. And we're reminded that at any given point, the potential is the combined effect of the potentials generated by all the charges present. So what our diagram shows is that we have point M we have this positive charge, it has a diameter of 0.2 m and a charge of 20 nano columns. And this is ball A, OK. At point N we have ball B with a diameter of 0.5 m. A charge of 80 nano Coombs. And the distance between the centers of ball A and ball B is 0.8 m. We're given four answer choices for the potential difference all in kill volts option A negative 0.38 option B negative 0.12 option C positive 0.12 and option D positive 0.38. So let's think about how to calculate our potential. OK. So the potential difference delta V, in this case, we're being asked to calculate is VM minus VN. OK. So what is the potential at VM? What is the potential at VM? Well VM and the potential at point M is going to be made up of the potential at point M generated by A and the potential at point M generated by B? OK. That's a reminder that we're given at the end of the problem text. So VM is going to be V MA plus VMB or V MA is the potential at point M generated by ball A and VMB potential at point M generated by ball B. OK. Similarly, for VNVN is going to be equal to VN A plus VNB. OK. So we get the potential of that point and generated by ball A and ball B. All right. So if we can calculate all four of these values V MA VMBV NAV NB, we'll be able to calculate our potential difference. Now, we're dealing with spherical balls here. So when we're talking about our potential, we wanna think about the potential of a sphere and that's gonna be one divided by four pi epsilon knot multiplied by Q divided by R. OK. So we know the charge of our balls, we are given all sorts of distances. So we'll be able to calculate R one divided by four pi epsilon knot. That's a constant value that we know. So we just need to go step by step to calculate those four values and solve for a potential difference. All right. So let's do it. And we are gonna start with all or with point M. OK? So we're gonna start with point M and at point M we're gonna start with the potential due to ball A. No, this is going to be equal to one divided by four pi epsilon knot multiplied by the charge of ball A divided by the distance between point M and the center of A. So this is gonna be R ma. All right. Now, the first part is easy. This is our constant value. So this is gonna be nine multiplied by 10 to the exponent nine Newton meters squared or Coolum squared. OK. The charge of um A we're given is 20 nano coulombs. So we're gonna take our 20 nano coulombs. We're gonna multiply by 10 to the exponent negative nine to convert this into coulombs. OK? And then we're gonna divide by the distance. Now, this distance again is the distance from the center of ball A to point A point M lies on the outside of Bale ball A has a diameter of 0.2 m. OK. That means it has a radius of 0.1 m half of that diameter. And so the distance from the edge of that ball where point M is to the center of the ball is going to be that radius of 0.1 m. OK? All right. So we're dividing by 0.1 m. Now, when we work this all out, we get a V MA value of 1800 volts. So in terms of our units here, we have Newton meter squared per Coolum squared. We're multiplying by coulombs dividing by meters. This leaves us with Newton meters per Coulomb. Ok. A Newton meter recall is equivalent to a jewel. So we have jewels per Coolum and that is equivalent to a volt. OK. So that's how we get from this kind of mess of units that we have into that nice clean unit of a vault. All right. So we're done V MA now we're gonna switch over to VMB. So again, we're still looking at point M, but now we're looking at the potential caused by BB at point M. So VMB, it's going to be equal to one divided by four pi epsilon not multiplied by QB divided by RMB substituting in our values. Again, that constant nine multiplied by 10 to the exponent nine Newton meters squared per Coulomb squared multiplied by the charge of B. It's 80 nano Coulombs multiplying by 10 to the exponent negative nine to convert to Coulombs. And then we're gonna divide by the distance between the center of ball B and point M. Now we're told that the center of ball B and the center of ball A are 0.8 m apart. OK? No point M sits on the edge of ball A closest to ball B. OK. So the distance between the center of B and point M is going to be that 0.8 m minus the radius of ball A which is 0.1 m. OK. So we have that total distance between the centers but we have to subtract the radius of ball A because M is on its surface came just a little bit closer to ball B. All right. Now, we can work all of this out. The units are gonna work out the exact same way. And what we get is 1028.57143 volts approximately. OK. We're gonna try to keep a decent amount of significant digits there so we can avoid some round off error. So now that we have V MA and V MD, we can calculate VM. OK? And that's the first piece of this puzzle for our potential difference. So VM is going to be the sum of the two. OK. The potential at point F is the potential from A plus, the potential from B at that point. So it's gonna be V MA plus VMB. When we add those two values together 1800 volts plus 1028.57143 volts. We get 2828.57143 volts. All right. So we're gonna put a red box around this. We don't wanna lose track of this value because we're gonna need it to do our final calculation. And now we're gonna switch over to point N in point N. We're doing the exact same process. OK. So we're looking at the potential at point N caused by ballet and then caused by bul, so let's go ahead and do it. So the potential at point N caused by A VN A is going to be one divided by four pi epsilon knot multiplied by Q A divided by RN A. OK. All right. So we have that constant value nine multiplied by 10 to the exponent nine Newton meters squared per Coulomb squared. The charge of A 20 multiplied by 10 to the exponent negative nine coulombs just like before. And now we have to figure out the distance between point N and the center of ball A. So let's go to our diagram and remember that the center of ball A and ball B are 0.8 m apart. Hm. Now the center of ball A to point N is gonna be a little bit closer because N sits on the surface of ball B closest to point or closest to ball A. So we have that total distance to their centers and then we have to subtract the radius of ball B. Ball B has a diameter of 0.5 m. So the radius is half of that 0.25 m. And so the distance between point N and the center of ball A is going to be 0.8 m minus 0.25 m. All right. Now, we work all of this out on our calculator and we get 327.27 volts approximately. That's the potential at point N caused by ball A. Now, we do the same for bulby potential at point N caused by BVNB one divided by four pi epsilon not multiplied by QB divided by R and B substituting in our values nine multiplied by 10 to the exponent nine Newton meters squared per coulomb squared multiplied by the charge of B 80 multiplied by 10 to the exponent negative nine coulombs divided by the distance between the center of B and point N. Back to our diagram one more time. OK. N sits at the surface of ball B and so its distance to the center of B is just going to be the radius of B which we've already determined to be 0.25 m. OK. So the distance between N and the center of B is going to be 0.25 m. And when we work this one out on our calculator, we get a potential of 2880 volts. OK. So we have now the potential at point N caused by A and caused by B, we can calculate the total potential at point N which is going to be the sum. OK. So the potential at N caused by A plus, the potential at N caused by B this is gonna be 327.27 repeated volts plus 2880 volts. Oops, which gives us a final value of 3207.27 repeated volts. All right. So we're putting a red box around that one as well. And now we can finally calculate what we wanted to do. OK. So this has been a long question, but we just needed to work through piece by piece. OK? So now we can calculate that potential difference that we are looking for. And the potential difference we were asked to find delta V was VM minus VN. So we're gonna take these two values we found and subtract them 2828.57143 volts minus 3207.27 volts. And what we get is a potential difference of negative 378 0.7013 volts. Now the question gave us the answer choices all in Koltz. So we're gonna have to convert this to Koltz to get our final answer to do that. We're gonna divide by 1000 and we get that the potential difference VM minus VN is going to be negative 0.3787 K volts. OK? And that is that final answer we were looking for. So going back to our answer choices comparing what we found. We're gonna round to two significant digits and we can see that the correct answer is going to be option a negative 0.38 kilowatts. Thanks everyone for watching. I hope this video helped see you in the next one.

FIGURE P25.67 shows two uniformly charged spheres. What is the potential difference between points 1 and 2? Which point is at the higher potential? Hint: The potential at any point is the superposition of the potentials due to all charges.

Key Concepts

Electric Potential

Superposition Principle

Potential Difference

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