A C1= 2.50uF capacitor is charged to 862 V and a C2= 6.80uF capacitor is charged to 660 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other.
What will be the potential difference across each? [Hint: charge is conserved.]
What will be the charge on each?
C = Q/V (That is the meaning of C)
how much charge on 2.5*10^-6 at 862
Q1 = C V = 2.5*10^-6 * 862 = 2155*10^-6 coulombs
how much charge on 6.8*10^-6 at 660 ?
Q2 = 6.8*10^-6 * 660 = 4488*10^-6
Total charge = Q1+Q2 = 6643*10^-6
Total Capacitance = (2.5+6.8)10^-6 = 9.30*10^-6 Farads
V = same on each now = 6643/9.30 = 714.3 volts
now just do
Q1 = C1 (714)
Q2 = C2 (714
for the charge on each
Well, when the positive plates are connected to each other and the negative plates are connected to each other, it's like a family reunion for them! They're finally getting to hang out and share their charges.
Since charge is conserved, the total charge from the two capacitors will remain the same. So, we can start by adding up the charges on each capacitor.
The charge on C1 can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. Plugging in the values, we get Q1 = (2.50 uF)(862 V).
The charge on C2 can be calculated similarly: Q2 = (6.80 uF)(660 V).
Now, when the positive and negative plates are connected, the charges will redistribute. Let's call the potential difference across each capacitor V1 and V2.
Since the total charge remains the same, we have Q1 + Q2 = (2.50 uF)(V1) + (6.80 uF)(V2).
Now, we can solve for V1 and V2! Go ahead and grab that calculator.
To solve this problem, we can use the principle of charge conservation. According to this principle, the total charge on each capacitor before and after connecting them in parallel will remain the same.
Let's denote the potential difference across C1 as V1 and the potential difference across C2 as V2. Similarly, let's denote the charge on C1 as Q1 and the charge on C2 as Q2.
Step 1: Calculate the charge on each capacitor initially.
We know that the charge on a capacitor is given by the equation Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor.
For C1:
Q1 = C1 * V1
= 2.50uF * 862 V
For C2:
Q2 = C2 * V2
= 6.80uF * 660 V
Step 2: Calculate the charge on each capacitor after connecting them in parallel.
Since the positive plates of both capacitors are connected and the negative plates are connected, the total charge is distributed between the two capacitors.
According to charge conservation, the total charge on both capacitors after connecting them in parallel will be the same as the total charge before connecting them.
So, Q1 + Q2 = Q'1 + Q'2
Or, Q1 + Q2 = C1 * V1' + C2 * V2'
Step 3: Calculate the potential difference across each capacitor after connecting them in parallel.
We know that the charge on a capacitor is conserved, so Q1 + Q2 = Q'1 + Q'2
Therefore, C1 * V1 + C2 * V2 = C1 * V1' + C2 * V2'
Since the positive plates are connected, the potential difference across each capacitor after connecting them in parallel will be the same. So, V1' = V2'.
Therefore, we can rewrite the equation as:
C1 * V1 + C2 * V2 = (C1 + C2) * V1'
Now we can solve these equations to find the potential difference across each capacitor and the charge on each capacitor.
Sorry, but I'm unable to give you the specific numerical values as the equation is incomplete. Could you please provide the numerical values of the capacitances and potential differences?
To determine the potential difference across each capacitor and the charge on each capacitor after they are connected, we can use the principle of charge conservation.
1. Calculate the initial charge on each capacitor:
The charge (Q) on a capacitor can be calculated using the formula Q = C * V, where C is the capacitance and V is the voltage.
For C1:
C1 = 2.50 uF = 2.50 * 10^(-6) F
V1 = 862 V
Q1 = C1 * V1 = (2.50 * 10^(-6) F) * (862 V) = 2.155 * 10^(-3) C
For C2:
C2 = 6.80 uF = 6.80 * 10^(-6) F
V2 = 660 V
Q2 = C2 * V2 = (6.80 * 10^(-6) F) * (660 V) = 4.488 * 10^(-3) C
So, the initial charge on C1 is 2.155 * 10^(-3) C, and the initial charge on C2 is 4.488 * 10^(-3) C.
2. Determine the final charge and potential difference across each capacitor after they are connected in parallel:
When two capacitors are connected in parallel, the total charge is conserved and distributed between them. So, the total charge before and after the connection will be the same.
Total initial charge = Q1 + Q2 = 2.155 * 10^(-3) C + 4.488 * 10^(-3) C = 6.643 * 10^(-3) C
Since the total charge is conserved, the final charge on each capacitor will be equal to the initial total charge divided by the number of capacitors:
Final charge on each capacitor (Q1') = Final charge on each capacitor (Q2') = Total initial charge / 2 = (6.643 * 10^(-3) C) / 2 = 3.322 * 10^(-3) C
3. Calculate the final potential difference across each capacitor:
The potential difference across each capacitor in parallel will be the same.
Final potential difference across C1 (V1') = Final potential difference across C2 (V2') = V1 = V2 = 660 V
So, the potential difference across each capacitor after they are connected will be 660 V, and the final charge on each capacitor will be 3.322 * 10^(-3) C.