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Lecture 8 - Two parallel plates Two parallel plates An...

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Two parallel plates An interesting observation. If the plates are large, they will acquire equal and opposite charge densities. - σ E= σ / ε 0 + σ
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Parallel Plates The potential difference between the two plates is then V + -V - = Ed= σ d/ ε 0 , where d is the plate separation. If we ground the negative plate we can write this as V= d/ ground the negative plate, we can write this as V= σ d/ ε 0 . If A is the area of the plates, we can identify the total charge on the positive surface as Q= σ A=( ε 0 A/d)V, and once more the charge on the plate is proportional to the voltage. The statement Q V, or Q=CV, with C a constant that depends only on geometry is true for any configuration of conductors. C tells us how much charge a given conductor can hold at a given voltage; i.e., the capacity , or in modern terms, the capacitance of the conductor. For the parallel plate capacitor C PP = ε 0 A/d. We also worked out last time C for an isolated sphere: worked out last time C for an isolated sphere: C S =4 πε 0 R, and C for a coaxial cable, C coax =2 πε 0 L/ln(b/a) .
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A coaxial cable Suppose V=V 0 on the inner conductor at r=a and V=0 on the outer conductor at r=b – a common choice. Let’s find the charge/length= λ for the cable. •Gauss’s Law Æ E= λ /2 πε 0 r. Gauss s Law E Then 0-V 0 =- λ dr/2 πε 0 r from r=a to r=b , or V 0 =( λ /2 πε 0 )ln( b/a ). N t th f l t f th l () f ti l ( b ) l ( ) Note the useful property of the ln() function: ln( )-ln( a ) = ln( b/a ). So λ =2 πε 0 V 0 /ln( b/a ). Notice that for a length L of the cable, Q= λ L is again proportional to V: Q= λ L=2 πε 0 L/ln( b/a )×V 0 = CV 0 And C coax =2 πε 0 L/ln( b/a ). 2
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Capacitance Units: from Q=CV, the units for C are C/V = Farad (F).
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