a 2uF capacitor is charged to a potential of 200v and then isolated. When it is connected in parallel switch a second capacitor which was initially not charged, the common potential becomes 40v. The capacitance of the second capacitor is?
To find the capacitance of the second capacitor, let's use the concept of charge conservation and voltage sharing in capacitors.
We start with the first capacitor, which is charged to a potential of 200V and has a capacitance of 2uF. Let's denote the charge on this capacitor as Q1.
The charge on a capacitor is given by the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential.
So, for the first capacitor: Q1 = (2uF)(200V) = 400uC
Now, when the first capacitor is connected in parallel with the second capacitor, charge will flow from the first capacitor to the second capacitor until the common potential between them reaches equilibrium. Let's denote the capacitance of the second capacitor as C2.
According to the concept of voltage sharing, the ratio of the voltage across each capacitor is inversely proportional to their capacitances. In other words:
V1/V2 = C2/C1
where V1 is the potential across the first capacitor (200V), V2 is the potential across the second capacitor (40V), C1 is the capacitance of the first capacitor (2uF), and C2 is the capacitance of the second capacitor (which we need to find).
Rearranging the formula, we have:
C2 = (V2/V1) * C1
Substituting the given values, we get:
C2 = (40V / 200V) * 2uF = (1/5) * 2uF = 0.4uF
Therefore, the capacitance of the second capacitor is 0.4uF.
To solve this problem, we can use the principle of charge conservation, which states that the total charge before and after the connection remains the same.
Let's denote the capacitance of the second capacitor as C2 and the charge on it as Q2.
Before the connection, the charge on the first capacitor, Q1, is given by:
Q1 = C1 * V1
where C1 is the capacitance of the first capacitor (2μF) and V1 is the potential on the first capacitor (200V).
After connecting the capacitors in parallel, the total charge becomes:
Q1 + Q2 = C1 * V1 + C2 * V2, where V2 is the potential on the second capacitor after the connection (40V).
Since the first capacitor was isolated and not connected to anything, its charge remains the same:
Q1 = C1 * V1
Substituting this back into the equation, we get:
C1 * V1 + Q2 = C1 * V1 + C2 * V2
Simplifying the equation, we find:
Q2 = C2 * V2
Since Q2 = C2 * V2, we can write:
C2 = Q2 / V2
Now, we need to find the value of Q2.
To find Q2, we can use the fact that the total charge before and after the connection remains the same:
Q1 = Q1 + Q2
Substituting the values, we have:
C1 * V1 = C1 * V1 + C2 * V2
Since the potential on the first capacitor is higher than the potential on the second capacitor, we can rearrange the equation to solve for C2:
C2 = (C1 * V1 - C1 * V1) / V2
C2 = 0 / V2
C2 = 0μF
Therefore, the capacitance of the second capacitor (C2) is 0μF.
energy stored = 1/2 * capacitance * [(voltage)^2]
capacitors in parallel equal the sum of the individual capacitances
1/2 * 2 * 200^2 = 1/2 * (2 + c) * 40^2
solve for c