When a switch is closed, an uncharged 151 nF capacitor is connected in a series circuit with a 10.0 x 106 ohm resistor and a 9.00 V battery. What is the charge on the capacitor 1.3 s after the switch is closed

Explanation:

Given that,

Capacitance of capacitor

C = 151 nF

Resistance of a resistor

R = 10×10^6 ohms

Battery EMF

V = 9V

Charge on capacitor after t = 1.3sec?

This is an RC series circuit

Current in the circuit is given as

i= dq/dt

dq = i•dt

q = ∫i•dt

The voltage in the capacitor is give as

q = CV

Vc = q/C, since q = ∫i•dt

Vc = 1/C ∫i•dt

Therefore applying KVL

Vr + Vc = V.

Vr is voltage across resistor = iR

iR + 1/C ∫i•dt = V

Note that, i=dq/dt and q = ∫i•dt

R dq/dt + q/C = V

Solving this differential equation

Divide through by R

dq/dt + q/RC = V/R

Since, R = 10×10^6 C = 151×10^-9F

Then, dq/dt + q/RC = V/R

dq/dt + q/10×10^6 × 151×10^-9= 9/10×10^6

dq/dt + 0.662q = 9×10^-7

Using integrating factor method

IF = e(0.662t)

So,

q•e(0.662t) = 9×10^-7∫(e(0.662t)dt

q•e(0.662t) = 9 × 10^-7•e(0.662t) / 0.662 + K

q•e(0.662t)=1.36×10^-7•e(0.662t) + K

Divide through by e(0.662t)

q = 1.36×10^-7 + K•e(-0.662t)

At the beginning, there was no charge on the capacitor

q(0) = 0

0 = 1.36 × 10^-7 + k

Then, k = -1.36 ×10^-7

q =1.36×10^-7—1.36×10^-7•e(-0.662t)

Now, at t = 1.3s

q=1.36×10^-7— 1.36 ×10^-7•e(-0.662 × 1.3)

q=1.36×10^-7-1.36×10^-7•e(-0.8606)

q =1.36×10^-7—5.75 × 10^-8

q = 7.85 × 10^-8

Then, q = 78.5 nC