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?

Question

Vikas TU · Answer

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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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?

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?

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