Question

20. [10pt] A RC series circuit, with R 6.50 and C-: 7.10 μ F, is connected through a switch to a battery with an emf of 5.50 Volt (assumed to have no internal resistance). If the capacitor is initially uncharged, what will the current through the resistor be 23.08 milliseconds ater the switch is closed? 21. [10pt] The electric potential at a point located 18.90 mm from a positive charge of 5.2 C and 70 mm from a second charge is measured to be 5.40 kV. Calculate the value of the second charge.

Answer 1

Q 20. time constant

τ = RC = (6.5X 10°)(7. 1 X 10-о) = 46.15X 10-3s = 46.15ms

Current in circuit at time t is given by

$i = \frac{E}{R} e^{^{-t/\tau}}$

Putting E = 5.5 Volt, R = 6.5 k$\Omega$, t = 23.08

We get

i = ​​​​​​0.513 mA

21.

T1

:q2

k T1 7
$\frac{q_{2}}{r_{2}}$

put V= 5400 volt, k= 9×10⁹, q₁= 5.2×10^-6, r₁= 18.9×10^-3 m, r₂ =23.7×10^-3 m

42.IS 6