The electric field between the two parallel plates of a capacitor of 1.5 μ F capacitance drops to one third of its initial value in 6.6 μ s when the plates are connected by a thin wire. The resistance of this wire is ________ Ω. (Given, log 3=1.1)

Solution

Physics sir · 4 steps

The field tracks the charge, so this is just RC discharge: find when the charge falls to one third.

1. Field follows charge. Between the plates E = q/(ε₀ A), so E is proportional to the charge q. If the field drops to one third, the charge has dropped to one third too.
2. Discharge law. Through the wire the capacitor discharges as q = q₀ e^(-t/RC). Put q = q₀/3 and take the natural log of both sides.

   1/3 = e^(-t/RC) ⇒ t = RC ln 3 ⇒ R = t/(C ln 3)

3. Put in numbers. The given log 3 = 1.1 is the natural log. R = (6.6 × 10⁻⁶)/(1.5 × 10⁻⁶ × 1.1) = 6.6/1.65.
4. So. R = 4 Ω, so the answer is 4.

The common slip: 4.4. Treating 6.6 μs as one time constant, t = RC, leaves out the ln 3 and gives 4.4 instead of 4.

The idea to keep. When a capacitor discharges, its charge, voltage and field all fall as e^(-t/RC). Falling to 1/n of the start takes t = RC ln n.

Answer from the NTA final answer key. Solution by PrepWiser.