Consider two isolated metal spheres,the first with a charge of 8.7 and a radius of 25 cm and the second uncharged with a radius of 5.4 cm.What is the charge on the smaller sphere after a long thin metal wire is used to connect the two spheres?

Question

Consider two isolated metal spheres,the first with a charge of 8.7  and a radius of 25 cm and the second uncharged with a radius of 5.4 cm.What is the charge on the smaller sphere after a long thin metal wire is used to connect the two spheres?

Quizplus · Accepted Answer

When the two spheres are connected by the wire, they will reach equilibrium and their charges will be distributed evenly. We can use the formula for the charge on a sphere, Q = kQr/R, where Qr is the charge on the sphere before they are connected, R is the radius of the sphere before they are connected, and k is a constant.

For the larger sphere:
Q1 = k(8.7 μC)(0.25m)/(0.25m) = 8.7 μC

For the smaller sphere:
Q2 = kQr/R = k(0)/(0.054m) = 0

So the total charge after they are connected is Q1 + Q2 = 8.7 μC. Since they are distributed evenly, the charge on the smaller sphere is 1.5 μC (the difference between the original charge on the larger sphere and the total charge after they are connected):

Q2 = Q1 - (Q1 + Q2) = -Q2 = -1.5 μC.

The negative sign just indicates that the charge on the smaller sphere is opposite in sign to the charge on the larger sphere.

Consider Two Isolated Metal Spheres,the First with a Charge of 8.7