Unformatted text preview: /ε0. You must find the
electric potential to find the capacitance ) 2) Find the equivalent capacitance of the above configuration. If the voltage across C1 is
3V, find the charge on C1 and C4. 1) From Gauss’s Law, the electric field is
E = λ/2πε0r
So the electric potential difference between the surfaces (r = a and r = b) is
V =(λ/2πε0)ln(b/a)
Going back to the definition of capacitance
C = Q/V
We plug in the voltage to get
C = 2πε0L/ln(b/a)
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 = 3VC1
Q2 = 3VC2...
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This note was uploaded on 01/22/2014 for the course PHYS 7D taught by Professor Barwick during the Spring '12 term at UC Irvine.
 Spring '12
 Barwick
 Capacitance

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