A 170 -pF capacitor is connected in series with an unknown capacitance, and as a series combination they are connected to a battery with an emf of 25.0 V. If the 170 -pF capacitor stores 135 pC of charge on its plates, what is the unknown capacitance?
Q1 = C1V1.
V1 = Q1/C1 = 135/170 = 0.794 Volts.
V2 = 25 - 0.794 = 24.206 Volts.
C2 = (0.794 / 24.206) * 170pF=5.576pF.
NOTE: The largest voltage appears across the smallest capacitor.
To find the unknown capacitance, we can use the equation relating the charge (Q) stored in a capacitor, the capacitance (C), and the voltage (V) applied to the capacitor:
Q = C * V
In this case, we know the charge (Q) stored in the 170-pF capacitor is 135 pC, and the voltage (V) applied across the capacitors is 25.0 V. Let's solve for the unknown capacitance (C).
C_unknown * V = Q
C_unknown = Q / V
Now we can substitute the values:
C_unknown = 135 pC / 25.0 V
To calculate capacitance in Farads (F), we need to convert pC to C:
1 F = 1 C
1 pC = 1 × 10^(-12) C
So, we have:
C_unknown = (135 × 10^(-12)) C / 25.0 V
Calculating this would give us the value of the unknown capacitance in Farads.
To find the unknown capacitance, we can use the formula Q = CV, where Q represents the charge stored, C represents the capacitance, and V represents the voltage.
Given information:
- Capacitance of the known capacitor (170 pF) = C1
- Charge stored on the known capacitor (135 pC) = Q1
- Capacitance of the unknown capacitor = C2
- Voltage (emf) = V (25.0 V)
In a series circuit, the charges stored on the capacitors are the same. Therefore, the charge on the unknown capacitor (Q2) will also be 135 pC.
Now we can use the formula Q = CV for both capacitors:
For the known capacitor:
Q1 = C1 * V
135 pC = 170 pF * 25.0 V
Solving for Q2:
Q2 = Q1 = 135 pC
Now we can find the unknown capacitance using the equation:
Q2 = C2 * V
135 pC = C2 * 25.0 V
To find C2, rearrange the equation:
C2 = Q2 / V
C2 = 135 pC / 25.0 V
Calculate the unknown capacitance:
C2 = 5.4 pF
Therefore, the unknown capacitance is 5.4 pF.