The electric field between the two parallel plates of a capacitor of 1.5 capacitance drops to one third of its initial value in 6.6 when the plates are connected by a thin wire. The resistance of this wire is . (Given, e 3 = 1.1)
Answer
4
Solution
E = Vd, the field decays like the voltage: E = E0e-t/RC
E03 = E0e-t/RC t = RC 3
R = tC 3 = 6.6 10-61.5 10-6 1.1
= 4
Answer: 4