Two capacitors, C1 = 27.0 µF and C2 = 35.0 µF, are connected in series, and a 21.0 V battery is connected across them.

(a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor.
equivalent capacitance=

µF
total energy stored=

(b) Find the energy stored in each individual capacitor.
energy stored in C1 J
energy stored in C2

The charge is the same on both

Q = C V
so C1V1 = C2 V2
V2 = (C1/C2)V1 and V1 + V2 = 21
let C1 = 27*10^-6
let C2 = 35*10^-6
then C1/C2 = .77
then V2 = .77 V1
1.77 V1 = 21
V1 = 11.9 volts
V2 = 9.1 volts

I think you can take it from there

To find the equivalent capacitance when capacitors are connected in series, you can use the formula:

1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...

In this case, there are only two capacitors connected in series, so the formula simplifies to:

1/Ceq = 1/C1 + 1/C2

Let's substitute the given values:
1/Ceq = 1/27 µF + 1/35 µF

To combine the fractions, you need to find a common denominator. The least common multiple of 27 and 35 is 945. So the equation becomes:
1/Ceq = (35/945 + 27/945) µF

Simplifying further:
1/Ceq = (62/945) µF

Now you can take the reciprocal of both sides to get the equivalent capacitance (Ceq):
Ceq = 945/62 µF ≈ 15.24 µF

So, the equivalent capacitance is approximately 15.24 µF.

To find the energy contained in the equivalent capacitor, you can use the formula:

E = (1/2) * C * V^2

where E is the energy, C is the capacitance, and V is the voltage.

Substituting the values:
E = (1/2) * 15.24 µF * (21.0 V)^2
E ≈ 3220.74 µJ

Therefore, the total energy stored in the equivalent capacitor is approximately 3220.74 µJ.

To find the energy stored in each individual capacitor, you can use the formula:

E = (1/2) * C * V^2

For C1:
E1 = (1/2) * 27.0 µF * (21.0 V)^2

Simplifying:
E1 ≈ 6172.5 µJ

Therefore, the energy stored in C1 is approximately 6172.5 µJ.

For C2:
E2 = (1/2) * 35.0 µF * (21.0 V)^2

Simplifying:
E2 ≈ 8082.75 µJ

Therefore, the energy stored in C2 is approximately 8082.75 µJ.