Question

A 12.6 μF capacitor is connected through a 0.890 MΩ resistor to a constant potential difference of 60.0 V. Compute the charge on the capacitor at the following times after the connections are made: 0, 5.00 s, 10.0 s, 20.0 s, and 100.0 s.

Answer 1

Answer:

a) 0 b) 272 μC . c) 446 μC d) 630 μC d) 756 μC

Explanation:

a) at t=0, as the voltage through a capacitor can't change instantanously, Vc must be 0.

By definition, the charge on a plate of a capacitor, the capacitance, and the voltage between the plates, are related by this expression:

C= Q/V, so if V=0, Q must be 0 also, due to C is a constant only dependent upon the geometry and the dielectric material between plates.

b) As current starts to flow, the charge begins to build upon the plates, generating a voltage that opposes to the applied voltage, decreasing the value of the current.

The charge on any plate of the capacitor, at any time, can be written as follows:

Q = CV (1- e⁻t/RC) so, at t= 5s, ⇒ Q= 272 μC

c) At t=10s, applying the same expression, we have:

Q= 446 μC

d) At t= 20s, using the same formula:

Q= 630 μC

e) When t reaches to 100 s, the capacitor is fully charged, reaching to the maximum possible value, CV, which is equal to 756 μC .

In this condition, no current remains flowing, as the capacitor has developed a voltage of the same value (and opposite sign) to the applied voltage.