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phyvamp
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Homework Statement
1)
Two solid conducting spheres have radius r1 = 3 cm and r2 = 42 cm. The two spheres are connected by a thin conducting wire. Assume:
- the wire is very long and thin, with negligible surface area, so it does not affect the capacitance of the system.
- the switch is closed.
- the spheres are very far apart, so they do not affect each other's charge distribution.
If charge Q is placed on sphere 1, it distributes between the two spheres. Find the capacitance of this system, in pF.
2)
Two solid conducting spheres have radius r1 = 6.98 cm and r2 = 30.9 cm. The two spheres are connected by a thin conducting wire, with the initially switch closed. Assume the wire is very long and thin, and the same assumptions apply. Charge 7.11 μC is placed on spheres, and distributes between the two spheres.
Now the radius of sphere 2 begins to decrease at the rate of:
dr2/dt = -2.54 cm/s.
Find dq1/dt the rate at which the charge on sphere 1 is changing, in μC/s. The sign indicates if the charge is increasing or decreasing.
Homework Equations
C=Q/V
The Attempt at a Solution
1) Since the switch is closed, the charge remains the same on sphere1 and sphere2?
C1 = Q/V = Q/(Q/4*pi*E0*r1) = 4*pi*E0*r1
C2 = Q/V = Q/(Q/4*pi*E0*r2) = 4*pi*E0*r2
then since it is in series Ctotal=1/(1/C1+1/C2)
Am I on the right approach?
2)Can anyone list steps that I can follow to solve it. I am really suck with integral rate
thank you for the help