A 5.0 uF capacitor is charged to a p.d of 10v. It is then removed from the supply and connected in parallel with an uncharged 20 uF capacitor. Calculate the charge on each capacitor after connection
(My working so far,
5×10^-6×10=5×10^-5C )I am stumped on what to do any further
To calculate the charge on each capacitor after connection, you can apply the principle of conservation of charge. The total charge before and after the connection remains the same.
Let's denote the charge on the 5.0 uF capacitor as Q1 and the charge on the 20 uF capacitor as Q2.
1. Initially, the 5.0 uF capacitor is charged to 10V. Therefore, Q1 = 5 × 10^-6 C (as you have correctly calculated).
2. When the capacitors are connected in parallel, they will share the total charge equally. The total charge is the sum of the individual charges on each capacitor.
Total charge = Q1 + Q2
Since Q1 = 5 × 10^-6 C, let's assume Q2 = x (the charge on the 20 uF capacitor).
Total charge = 5 × 10^-6 C + x
3. According to the principle of conservation of charge, the total charge remains the same. Therefore, the total charge after connection is equal to the total charge before connection.
Total charge after connection = Total charge before connection
5 × 10^-6 C + x = 5 × 10^-5 C
4. Now you can solve for x by subtracting 5 × 10^-6 C from both sides of the equation.
x = 5 × 10^-5 C - 5 × 10^-6 C
= 4 × 10^-5 C
Therefore, the charge on the 5.0 uF capacitor after connection is 5 × 10^-6 C, and the charge on the 20 uF capacitor after connection is 4 × 10^-5 C.
To calculate the charge on each capacitor after they are connected in parallel, you can use the principle of charge conservation. The total charge before and after the connection should be the same.
Let's calculate the initial charge on the 5.0 uF capacitor:
Q1 = C1 * V1
= 5.0 * 10^-6 F * 10 V
= 5.0 * 10^-5 C
The initial charge on the 5.0 uF capacitor is 5.0 * 10^-5 C.
Now, since the capacitors are connected in parallel, the total charge will be distributed between them. Let Q2 be the final charge on the 20 uF capacitor. According to the principle of charge conservation:
Q1 + Q2 = (C1 + C2) * V
Q2 = (C1 + C2) * V - Q1
In this case, C1 is the capacitance of the 5.0 uF capacitor, C2 is the capacitance of the 20 uF capacitor, and V is the voltage across both capacitors.
Plugging in the values:
Q2 = (5.0 * 10^-6 F + 20 * 10^-6 F) * 10 V - 5.0 * 10^-5 C
Simplifying:
Q2 = (25.0 * 10^-6 F) * 10 V - 5.0 * 10^-5 C
= 2.5 * 10^-4 C - 5.0 * 10^-5 C
= 2.0 * 10^-4 C
The final charge on the 20 uF capacitor is 2.0 * 10^-4 C.
Therefore, the charge on each capacitor after connection is:
- 5.0 uF capacitor: 5.0 * 10^-5 C
- 20 uF capacitor: 2.0 * 10^-4 C
charge stored: 5E-5 (CV)
now, when connected, the two capacitors have equal voltages..
V=Q1/C1 =Q2/C2
but Q1+Q2= original charge , or
Q1+Q2=(5E-6 *10)
q2=5E-5-Q1
V=Q1/C1 =Q2/C2
Q1=C1/C2 * (5E-5-Q1) and C1/C2=1/4
which leads to
5/4 Q1=1/4*(5E-5)
Q1=.... then solve for Q2