Re: [問題] jackson中的電容C
Q_1 = C_11 V_1 + C_21 V_2
Q_2 = C_12 V_1 + C_22 V_2
Now consider the usual case in which Q_1 = -Q_2 = Q ,
=>
C_11 V_1 + C_21 V_2 = Q
C_12 V_1 + C_22 V_2 = -Q
then by Cramer's rule,
V_1 = Q(C_21+C_22)/det(C_ij)
V_2 = -Q(C_11+C_12)/det(C_ij)
=> V = V_1-V_2 = Q Σ_ij C_ij / det(C_ij) = Q/C
thus the capacitance we have learned before is given by
C = det(C_ij)/Σ_ij C_ij
我猜chungweitw大的意思應該是這樣
※ 引述《mouwat (QQ)》之銘言：
: : 我有再看了課本 但不是很懂= =
: : 老師上課有講一個簡單例題
: : 半徑a跟b的同心圓殼 算出來的C11=-C12=-C21=/=C22
: : 最後就取C=C11 這怎麼去決定?
: : C11=ab/(b-a) C12=C21=-ab/(b-a) C22=b^2/(b-a)
: the coefficients Cij are capacities
: or capacitances while Cij,i=/=j,are called coefficients of induction.
: thecapacitance of a conductor is therefore the total charge on the conductor
: when it is maintained at unit potential,
: all other of conductors beingheld at zero potential.
: sometimes the capacitance of a system of conductors is also defined.
: for example,
: the capacitance of two conductors carrying equal and opposite charges
: in the presence of other grounded conductors is defined as the ratio
: of the charge on one conductor to the potential difference between them.
: the equations Qi=summation CijVi can be used to express this capacitance
: in terms of the coefficients Cij
: 如果沒錯你應該是說這段吧
: 可是除了提到V=V1-V2
: 我不太了解哪裡敘述可以導出你提到的C=det(Cij)/sum(Cij)
: 是我英文不好嗎~"~
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◆ From: 140.112.213.158
※ 編輯: mantour 來自: 140.112.213.158 (11/18 22:11)
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11/18 23:08, , 1F
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