Unformatted text preview: find the
electric field between the two spheres). 2) Find the equivalent capacitance of the above configuration. If the voltage across C1 is
V, find the charge on C1 and C4. 1) From Gauss’s Law, the electric field is
E = KQ/r2
So the electric potential difference between the surfaces (r = a and r = b) is
V = KQ(1/b
1/a)
Going back to the definition of capacitance
C = Q/V
We plug in the voltage to get
C = ab/(k(a
b))
2) Remembering the rules for adding capacitors:
Series: C23 = (1/C2 + 1/C3)^
1 = C2C3/(C2+C3)
Parallel: Cnet = C1 + C23 + C4 = C1 + C4 + C2C3/(C2+C3)
To find the charge on C1, we go back to the definition of capacitance.
C = Q/V → Q = VC
So the charge on each capacitor is just
Q1 = VC1
Q2 = VC2 Quiz  Week 5  Capacitors 1) A cylindrical conductor has line charge λ. It is surrounded by a hollow conducting cylinder of
line charge 
λ. The region in between the two cylinders (a < r < b) is empty space.
i) Find the capacitance of the above configuration (hint: The total charge on the inner
cylinder is Q = λ*L. To find the electric field, we have E(2πr) = λ*L...
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 Spring '12
 Barwick
 Capacitance, Energy, Electric charge, Inductor, cnet

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