1cheers yaar!!
In the following R-C circuit, the capacitor is in steady state. The initial separation b/w the capactor plates is xo. If at t=0, the separation between the plates starts changing so that a constant current starts flowing through R, find the velocity with which the plates are moving.
P.S.: the plate area is A.
Figure:
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4 Answers
49E = qC + iR = qxeoA + iR..(i) (when plate separation is x, C= eoAx)
iR const E is const!
so differentiating w.r.t. time we get,
0=qeoA(dxdt) + xeoA(dqdt)
or, 0 = qveoA + ixeoA
giving, q= - ixv...(ii)
putting in eq (i)
E= - ix2eoAv + iR
or, -v= (i/eoAE-iR)x2
or, -dxdt= (i/eoAE-iR)x2
or, -dxx2= (i/eoAE-iR)dt
integrating with proper limits,
1x-1xo= (i/eoAE-iR)t ...(iii)
also, v= - (i/eoAE-iR)x2 =(i/eoAiR-E)x2...(iv)
from (iii) and (iv)
v= (i/eoAiR-E)x2)[(i/eoAE-iR)t + 1xo]
hence done [1][1]
was a nice MATHS question :P