Two capacitor 3.0 pico Farad and 6.0 pico Farad are connected in series. If the capacitor connected to a potential difference of 1000v, calculate
a. Magnitude of the charge on each capacitor
b. The potential difference across the capacitor
c = q/v
same q on each, call it q
so
c1 V1 = c2 V2 so V2 = (c1/c2) V1
and
V1 + V2 = 1000
V1 + (c1/c2)V1 = 1000
but c1/c2 = 3/6 = .5
so
1.5 V1 = 1000
V1 = 667
V2 = 333
I think you can go back and get q = v c for each
To solve this problem, we can use the properties of capacitors connected in series.
a. To find the magnitude of the charge on each capacitor, we need to use the formula:
Q = CV,
where Q is the charge, C is the capacitance, and V is the potential difference applied to the capacitor.
For the first capacitor with a capacitance of 3.0 pico Farad (pF) and a potential difference of 1000 V, we can calculate the charge as follows:
Q1 = (3.0 pF) * (1000 V) = 3000 pico Coulombs (pC).
Similarly, for the second capacitor with a capacitance of 6.0 pF and the same potential difference of 1000 V:
Q2 = (6.0 pF) * (1000 V) = 6000 pico Coulombs (pC).
Therefore, the magnitude of the charge on each capacitor is 3000 pC for the 3.0 pF capacitor and 6000 pC for the 6.0 pF capacitor.
b. The potential difference across each capacitor in a series connection is divided proportionally based on the capacitance values. To find the potential difference across each capacitor, we can use the formula:
V = Q / C,
where V is the potential difference, Q is the charge on the capacitor, and C is the capacitance.
For the first capacitor with a charge of 3000 pC and a capacitance of 3.0 pF:
V1 = (3000 pC) / (3.0 pF) = 1000 V.
Similarly, for the second capacitor with a charge of 6000 pC and a capacitance of 6.0 pF:
V2 = (6000 pC) / (6.0 pF) = 1000 V.
Therefore, the potential difference across each capacitor in this series circuit is 1000 V.