1that is to be proved mani!
Take a system of two conductors of any shape and size with charges Q1 and Q2 and potentials V1 and V2.Let q be the charge that would flow from the conductor when they are shorted.Prove that the quantity q/(V1-V2) is a constant depending only on the geometry of the configuration.This quantity may be interpreted as the capacitance of the system.Does it reduce to the usual definition when Q1=-Q2 .
Please help me with this one ... I have my method but that is really doubtful ..
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8 Answers
11bro as far as i know the capacitance of the system depends upon the shape of the capacitor
1Let us define a quantity f_{1} depending on the geometry of body 1 such that if the body 1 has charge Q_{1} then the potential of body due to itself is Q_{1}f_{1}
Similarly we have f_{2}
let G_{12}=G_{21}=G be a similar quantity describing the effect of one body on the other.
so, we have :
V_{1}=Q_{1}f_{1}+Q_{1}G
V_{2}=Q_{2}f_{2}+Q_{2}G
let q is the charge flown when they are shorted.
so, we have:
(Q_{1}-q)f_{2}+(Q_{1}-q)G =(Q_{2}+q)f_{2}+(Q_{2}+q)G
so, q=\frac{(Q_{1}-Q_{2})f_{2}+(q_{1}-Q_{2})G}{f_{1}+f_{2}+2G}
so, we get that \frac{q}{V_{1}-V_{2}}=f_{1}+f_{2}+2G
which is dependent only on the geometry of the two bodies.
1well even if I don't wok with that piece of i information I think I will still get the proof!
Then I will get :\frac{q}{V_{1}-V_{2}}=f_{1}+f_{2}+G_{12}+G_{21}
1yes i was also wondering abt that . I thought it to be a function of four variables