A capacitor A1B1 of capacitance C1=2microF is charged under a voltage U1=20V and then isolated.
a) Calculate the voltage U (A1B1) between its plates, and it's charge Q1 and energy W1.
b) The plates A1 and B1 are connected to those A2 and B2 of an uncharged capacitor of capacitance C2=8microF. At electric equilibrium the respective charges of the capacitor becomes Q'1 and Q'2. Establish a relation between Q1, Q'1 and Q'2.
Compare the final energy of the system to the initial energy of the capacitor (A1B1).
P.S:I only need part b.
Thank you
Use the work-energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases.
I think you're mistaken
Since the second capacitor is 4x as big it will hold charge in a 4:1 ratio.
To answer part b of your question, let's analyze the situation after connecting the plates A1 and B1 of the first capacitor to those A2 and B2 of the second capacitor.
When the capacitors are connected, charge redistribution occurs until the system reaches electrical equilibrium. At this point, the final charges on the capacitors will be denoted as Q'1 and Q'2.
To establish a relation between Q1, Q'1, and Q'2, we can use the principle of charge conservation. According to this principle, the total charge before and after connecting the capacitors must be the same.
Initially, capacitor A1B1 has a charge of Q1. Since it was isolated after being charged, this charge remains constant.
When the capacitors are connected, the total charge is redistributed between the capacitors until reaching equilibrium. The total charge of the system is given by the sum of the individual charges on each capacitor:
Total charge = Q1 + Q'2
Since the total charge remains constant, we can express Q'1 in terms of Q1 and Q'2 as:
Q'1 = Total charge - Q'2
= (Q1 + Q'2) - Q'2
= Q1
Therefore, the relation between Q1, Q'1, and Q'2 is simply:
Q'1 = Q1
This shows that the charge on the first capacitor, A1B1, remains unchanged after connecting it to the second capacitor.
To compare the final energy of the system to the initial energy of the first capacitor, we need to consider the energy stored in each capacitor before and after connection.
The energy stored in a capacitor is given by the formula:
Energy = (1/2) * Capacitance * Voltage^2
Initially, the energy stored in capacitor A1B1 (W1) can be calculated using the given values of capacitance (C1) and voltage (U1):
W1 = (1/2) * C1 * U1^2
After connection, the final energy of the system (Wf) is the sum of the energies stored in both capacitors:
Wf = (1/2) * C1 * U^2 + (1/2) * C2 * U^2
However, we need to determine the voltage (U) across the plates A1B1 and A2B2. To find this, we can use the fact that the charge on both capacitors in electrical equilibrium is the same, i.e., Q'1 = Q'2.
Since Q'1 = Q1, we can express Q'2 in terms of Q1:
Q'2 = Q1
The capacitance of the second capacitor is given as C2, and we know that Q'2 = C2 * U. Substituting the values, we get:
C2 * U = Q1
Solving for U:
U = Q1 / C2
Now, we can substitute this value of U in the expression for final energy:
Wf = (1/2) * C1 * (Q1 / C2)^2 + (1/2) * C2 * (Q1 / C2)^2
Simplify the expression:
Wf = (1/2) * ((C1 * Q1^2) / C2 + (C2 * Q1^2) / C2)
Wf = (1/2) * (C1 + C2) * Q1^2 / C2
Comparing this expression with the initial energy of the first capacitor (W1), we can see that:
W1 / Wf = (C2 / (C1 + C2))
Therefore, the final energy of the system compared to the initial energy of capacitor A1B1 is given by the ratio of the capacitance of the second capacitor to the sum of the capacitance of both capacitors:
W1 / Wf = C2 / (C1 + C2)
Note: Make sure to double-check all calculations and units to ensure accuracy when working through the numerical values.