Question

1. In an RC circuit, the battery has a voltage of 12.0V and the R = 5 MΩ and C= 2.5μF. How long will it take in seconds for the charge to build up to 22.3 microcoulombs? (Show work please)

2. Richard has a 200 V battery and two metal disks, both with an area of 0.86 m^2 each and a distance
of 0.240 mm. How much charge in microCoulomb can his capacitor store? (please show work)

Answer 1

1. here we have given that,

voltage = 12.0V

R = 5 MΩ and

C= 2.5μF

SO THAT THE TOTAL CHARGE OVER THE CAPACITOR AT T = INFINITY WILL BE AS,

Qo = CV = 3 *10-5 C

so there time constant is given as

time constant T = RC = 12.5

Now here we can see that there is no initial charge on the capacitor so that required equation for the charge is as follows,

Q = Qo(1-e-t/T) ........................1

for the Q =  22.3 microcoulombs = 22.3 * 10-6 C

so that, from equation 1 we have,

22.3 * 10-6 =   3 *10-5 (1-e-t/12.5)

0.7433333 = 1-e-t/12.5

e-t/12.5 = 1- 0.7433333

e-t/12.5 = 0.2566666

nowhere on taking ln both sides we will get,

-t/12.5 = - 1.359977

therefore t = 16.9997 sec

hence capacitor will take  t = 16.9997 in seconds for the charge to build up to 22.3 microcoulombs.

2.

voltage = 200 V

area of 0.86 m^2

distance = 0.240 mm

epsilon not = 8.85* 10^-12 Nm2/c2

so here capacitance will be given as,

C = epsilon not *A / d =  8.85* 10^-12 * 0.86/ 0.240 *10^-3 = 3.17127 * 10^-8 C

so that ,  charge in microCoulomb can his capacitor store is given as,

Q = CV = 3.17127 * 10^-8 * 200 = 6.3425 * 10^-6 C =  6.3425 microCoulomb

which is our required answer.