Capacitor A has a capacitance CA =29.4 uF and us initially charged to a voltage of V1= 10.70 V. Capacitor B is initially uncharged . When the switches are closed, connecting the two capacitors, the voltage on capacitor A drops to V2 = 6.40 V. What is the capacitance CB (in uF) of capacitor B?
The voltage on A dropped from 10.70 to 6.40 V, or by 4.30 Volts. Capacitor A lost C(A)*deltaV = 1.264*10^-4 Coulombs of charge.
That charge flowed to capacitor B, leaving it with 1.264*10^-4 Coulombs and a charge of 6.40 V
C(B) = Q/V = 1.264*10^-4/6.4 = 19.8 uF
To find the capacitance CB of capacitor B, we can use the formula:
Q = CV
Where Q is the charge stored in the capacitor and C is the capacitance.
First, let's find the charge stored in capacitor A before and after the switch is closed.
The charge Q1 stored in capacitor A before the switch is closed is given by:
Q1 = CA * V1
Q1 = 29.4 uF * 10.70 V
Q1 = 314.58 uC
The charge Q2 stored in capacitor A after the switch is closed is given by:
Q2 = CA * V2
Q2 = 29.4 uF * 6.40 V
Q2 = 188.16 uC
Now, since capacitor B is initially uncharged, the charge transferred from capacitor A to capacitor B is equal to the difference in charge:
Q_transfer = Q1 - Q2
Q_transfer = 314.58 uC - 188.16 uC
Q_transfer = 126.42 uC
Now, we can use the formula Q = CV for capacitor B to find its capacitance CB:
Q_transfer = CB * V2
126.42 uC = CB * 6.40 V
CB = 126.42 uC / 6.40 V
CB ≈ 19.75 uF
Therefore, the capacitance CB of capacitor B is approximately 19.75 uF.
To find the capacitance (CB) of capacitor B, we can use the principle of conservation of charge.
Let's start by understanding the concept of capacitance and how it relates to voltage and charge.
The capacitance of a capacitor (C) is a measure of its ability to store charge. It is defined as the ratio of the charge (Q) stored on the capacitor to the voltage (V) across it:
C = Q / V
In this problem, we have two capacitors, A and B. Capacitor A has a capacitance of CA = 29.4 uF and is initially charged to a voltage of V1 = 10.70 V. Capacitor B is initially uncharged.
When the switches are closed, the charge on capacitor A redistributes, causing its voltage to drop to V2 = 6.40 V. At the same time, capacitor B becomes charged.
Since the total charge must remain constant, we can equate the charges before and after the switches are closed:
QA = QB
Using the formula for capacitance, we can express the charges as:
QA = CA * V1
QB = CB * V2
Substituting these equations into the equation QA = QB, we get:
CA * V1 = CB * V2
Now we can solve for CB:
CB = (CA * V1) / V2
Substituting the given values:
CB = (29.4 uF * 10.70 V) / 6.40 V
CB = 48.95 uF
Therefore, the capacitance of capacitor B (CB) is 48.95 uF.