practice problem 1
- a positively charged conducting sphere
- a positively charged conducting "pointy sphere"
- two oppositely charged, parallel conducting plates
practice problem 2
- the spheres are the same size
- the positive sphere has twice the radius of the negative sphere
With two spheres of equal size, the total charge will try to distribute itself evenly between them.The positive charges are repelled by one another and try to get as far apart as possible. The negative charges behave the same way. They best strategy for maximizing separation is to send half your members to one sphere and half to the other. Since neither sphere is more "attractive" than the other, the separation should be an even one. On the whole, we have…
Qnet = (+6 μC) + (−12 μC) = −6 μC
of charge that will separate evenly into to batches of…
Q1 = Q2 = −6 μC = −3 μC 2
If you were a charge, running away from the other charges and you had to decide between two conductors of different size, which would you choose? Well, you can't all choose the bigger one because then you'd be closer together than if some of you chose the big one and some the small one. Would you divide up evenly? No, that doesn't make much sense either. The small sphere would be pack tight and the big one would have unused capacity for holding charge. No, the "logical" conclusion in the minds of our anthropomorphic charges is to distribute themselves at arm's length against the outside of each sphere. The next thing to do is test the common wisdom with mathematics.
The quantity that determines the motion of charges is voltage — potential difference, that is. Charge will flow from high potential to low potential in much the same way that water flows from regions of high altitude to regions of low altitude. Our two charged spheres are like two lakes. Putting the spheres in contact is like connecting two lakes with an aqueduct. The higher one (higher in altitude) will drain away into the lower one until both lakes are at the same level. Similarly, our two charged spheres will exchange charge until they both reach the same voltage.
V1 = V2 ⇒ kq1 = kq2 ⇒ q1 = q2 r1 r2 2r r
At the same time, charge must be conserved. We started with −6 μC overall and so we shall end. Mathematically, we write this as…
q1 + q2 = −6 μC ⇒ q2 = −6 μC − q1
All that remains is to combine the two equations and solve for the two unknowns.
q1 = −6 μC − q1 2r r q1 = −12 μC − 2q1 3q1 = −12 μC q1 = −4 μC q2 = −2 μC
Well whad'ya know. The one that's twice as big (has twice the radius) can hold twice as much charge. That makes sense.
practice problem 3
practice problem 4