750_Physics ProblemsTechnical Physics

750_Physics ProblemsTechnical Physics - 90 Capacitance and...

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90 Capacitance and Dielectrics *P26.42 (a) The total energy is UU U q C q C q R Qq R =+= + = + 12 1 2 1 2 2 2 1 2 01 1 2 0 2 1 2 1 2 1 24 1 2 4 π bg . For a minimum we set dU dq 1 0 = : 1 2 2 4 1 2 2 4 10 1 1 02 21 1 11 1 1 q R R Rq RQ Rq q RQ RR ππ + −= =− = + a f Then qQ q q 2 2 =− = + = . (b) V kq R kRQ RR R kQ e ee 1 1 1 1 11 2 1 2 == + = + V R e 2 2 2 2 21 2 + = + and VV 0 . Section 26.5 Capacitors with Dielectrics P26.43 (a) C A d = = ×× × = −− κ 0 12 4 5 11 2 10 8 85 10 1 75 10 400 10 813 10 813 .. . . Fm m m F p F 2 ej e j (b) VE d max max . × × = 60 0 10 4 00 10 2 40 65 Vm m kV e j P26.44 QC V max max =∆ , but d max max = . Also, C A d = 0 . Thus, Q A d Ed A E max max max = =∈ 0 0 . (a) With air between the plates, = 100 . and E max . 300 10 6 Vm. Therefore, QA E max max . . = × × × = 0 12 4
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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