1anyone??
A spherical capacitor has inner and outer radius a,b
respectively.The space is suddenly filed with a medium of dielectric constant k=1.The charge on
inner plate is +q,outer plate -q.Suddenly,the medium becomes conducting with conductivity Ï.
taking this instant as t=0,determine the charge remaining on +ve plate at time t.
Ans.qe-tÏ/e0
-
UP 0 DOWN 0 1 3
3 Answers
11let at any instant charge on inner plt is q1 and in outer is q2.
so the pd between two plates is
(KQ1/A+KQ2/B)-(KQ2/B+KQ1/B)=KQ1(1/A-1/B)
v=IR
=>I=V/R
FOR FINDING R LET TAKE A CELL OF RADIOUS X AND THICKNESS dX
SO RESISTANCE ACROSS THIS IS dR=1/Ï*dX/4Ï€X2
INTEGRATING BOTH SIDES WITH SUITABLE LIMITS
=>R=1/4Ï€Ï(1/B-1/A)
SO FROM THIS I=-KQ1*4Ï€Ï
=>dQ1/dt=-MQ1(M=K4Ï€Ï TAKEN)
=>dQ1/Q1=-Mdt
INTEGRATING BOTH SIDES
=>ln(Q1/q)=-Mt=-t/4πξ *4Ï€Ï=-tÏ/ξ
=>Q1=-qetÏ/ξ