If [tex]$C_1 = C_2 = 4.00 \mu F$[/tex] and [tex]$C_4 = 8.00 \mu F$[/tex], what must the capacitance [tex]$C_3$[/tex] be if the network is to store [tex]2.30 \times 10^{-3} J[/tex] of electrical energy?

To determine the capacitance C3 in a network that must store 2.30×10−3J of electrical energy, we need to find the equivalent capacitance of the network and then solve for the voltage across this network. From there, we can calculate C3 using the series and parallel capacitance formulas, considering C1 and C2 in series and the result in series with C3, which is in parallel with C4.

Explanation:

To find the capacitance C3 needed for a capacitor network to store 2.30×10−3 J of energy, we will use the formula for the energy stored in a capacitor network, which is:

U = (1/2)CeqV2

However, we first need to find the equivalent capacitance Ceq of the network. If capacitors C1 and C2 are in series with a capacitor C3, and this combination is in parallel to C4, then

Ceq = Cparallel + Cseries

Given that C1 and C2 are the same and in series:

Cseries = C1/2

And the equivalent of Cseries and C3 in parallel with C4 is:

Cparallel = C4

Since energy U and C4 are already given:

Cseries + Cparallel = (1/2)C1 + C4

Ceq = (1/2)×4.00 μF + 8.00 μF

We can solve for the voltage across the network V using the energy formula:

V = √(2U/Ceq)

With the voltage and the known capacitances, we can then find C3 since C1, C2, and C3 are in series and their equivalent capacitance is in parallel with C4.

The necessary capacitance C3 allows the network to store the given energy with the calculated voltage from Ceq.