Answer:
[tex]t=14.678\times 10^{-3}s[/tex]
Explanation:
Given:
Capacitance, C = 85 pF = 85 × 10⁻¹² F
Resistance, R = 75 MΩ = 75×10⁶Ω
Charge in capacitor at any time 't' is given as:
[tex]Q=Q_o(1-e^{-\frac{t}{RC}})[/tex]
where,
Q₀ = Maximum charge = CE
E = Initial voltage
t = time
also, Q = CV
V= Final voltage = 90% of E = 0.9E
thus, we have
[tex]C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})[/tex]
or
[tex]0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}[/tex]
or
[tex]e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1[/tex]
taking log both sides, we get
[tex]ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})[/tex]
or
[tex]-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)[/tex]
or
[tex]t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}[/tex]
or
[tex]t=14.678\times 10^{-3}s[/tex]
