The drawing shows two fully charged capacitors (C1 = 7 µF, q1 = 4 µC; C2 = 16 µF, q2 = 4 µC). The switch is closed, and charge flows until equilibrium is reestablished (i.e., until both capacitors have the same voltage across their plates). Find the resulting voltage across either capacitor

you have 8microC total charge.

You have 23microF capacitance.

Voltage= Qtotal/C=8/23 volts.

on C1, q=C1*8/23=7micro*8/23=56/23 micro F
on C2, q=16micro*8/23=128/23 microC

adding q1+q2, 184/23 microF=8microF

Depending on which way they are wired:assume they are wired positive plate to negative plate of each other.

Then Qeff=16*4-7*4=36uC.(If wired positive plate to positive plate then charges will add up)

This charge will get redistributed.

Applying conservation of charge:

Qeff=(C1+C2)Vfinal

Vfinal=1.56v

The first answer is correct...final voltage will be: Qtotal/(C1+C2)=0.347

To find the resulting voltage across either capacitor, we need to use the concept of charge conservation and the relationship between capacitance, charge, and voltage.

The charge on a capacitor is given by the formula:

q = C * V

Where q is the charge, C is the capacitance, and V is the voltage across the capacitor.

In this scenario, we are given the charges on the capacitors (q1 = 4 µC, q2 = 4 µC) and the capacitances (C1 = 7 µF, C2 = 16 µF).

Initially, the capacitors are fully charged, so we can calculate the initial voltages using the formula:

V1 = q1 / C1
V2 = q2 / C2

Substituting the given values:

V1 = (4 µC) / (7 µF) ≈ 0.57 V
V2 = (4 µC) / (16 µF) ≈ 0.25 V

When the switch is closed, charge will flow between the capacitors until they have the same voltage across their plates. Let's denote this final voltage as V.

Now, charge is conserved, so the total charge on both capacitors remains the same:

q1 + q2 = (C1 * V) + (C2 * V)

Substituting the given values:

(4 µC) + (4 µC) = (7 µF * V) + (16 µF * V)

8 µC = 23 µF * V

Simplifying the equation:

V = 8 µC / 23 µF ≈ 0.35 V

Therefore, the resulting voltage across either capacitor is approximately 0.35 V.