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2 metal balls of rad r are placed at very large distance from each other. they are connercted by a coil of inductance l and a switch s. one sphere is given charge +q and switch is closed. after what time charge on sphere is reduced to half of its initial value?

Ambarish Kumar Banerjee , 8 Years ago
Grade 12th pass
anser 1 Answers
Vikas TU

Last Activity: 8 Years ago

As the whole system is placed at ground the circuit becomes closed loop and hence from KVL,
V – Ldi/dt – V’ = 0
V = Ldi/dt + V’
kq/r = Ldi/dt + 0 (initially)
kq/r = L*d^2q/dt^2
Ld^2q = (kq/rL)dt^2
qLdq = (kqt/rL)dt 
q^2*L = k*q*t^2/r*L
t^2 = qrL^2/k
t = root(qrL/k)
put q = q/2
t = root(qrL/2k) s 
 

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