Two capacitors, C1=25.0x10^-6 and C2=5.0x10^-6 are connected in parallel and charged with a 100-V power supply.

a)calculate the total energy stored in the 2 capacitors
b) what potential difference would be required across the same 2 capacitors connected in series in order that the combination store the same energy as in a)?

ENERGY STORED=1/2C*DELTA V^2
1/2(25.0e^-6 + 5.0e^-6)(100)^2
=0.15KJ OR 150J, IS THIS RIGHT? IF NOT HOW DO I GET THE RIGHT ANSWER?

B)1/Ceq= 1/25.0e^-6 + 1/5.0e^-6= 4.16e^-6
DELTA V= SQRT 2x (ENERGY STORED=150J)/
4.16e^-6 = 8492v?????? CAN SOMEONE TELL ME IF IM DOING THIS RIGHT?

The stored energy that you calculate with that formula is in J, not kJ. The answer to the first part is 0.15 J

For the second part, require that
(1/2)Ceq*V^2 = 0.15 J, and solve for V. You are correct that Ceq = 4.167*10^-6 F

so for part two i get 268V is that correct?

Yes

yes

To calculate the total energy stored in the two capacitors connected in parallel, you are correct in using the formula:

Energy = 1/2 * (C1 + C2) * (ΔV)^2

Plugging in the given values:

C1 = 25.0 x 10^-6 F
C2 = 5.0 x 10^-6 F
ΔV = 100 V

Energy = 1/2 * (25.0 x 10^-6 + 5.0 x 10^-6) * (100)^2
= 1/2 * 30.0 x 10^-6 * 10,000
= 1.5 x 10^-3 J
= 1.5 mJ

So the correct answer for part a) is 1.5 mJ.

Now, for part b), to find the potential difference required across the two capacitors connected in series to store the same energy:

First, you need to calculate the equivalent capacitance (Ceq) for capacitors connected in series. The formula is:

1/Ceq = 1/C1 + 1/C2

Plugging in the given values:

1/Ceq = 1/(25.0 x 10^-6) + 1/(5.0 x 10^-6)
= 40.0 x 10^6 + 200.0 x 10^6
= 240.0 x 10^6

Ceq = 1 / (240.0 x 10^6)
= 4.17 x 10^-6 F

Now, to calculate the potential difference (ΔV) required:

ΔV = √(2 * Energy / Ceq)
= √(2 * 1.5 x 10^-3 / (4.17 x 10^-6)
= √(0.72 x 10^3)
= √(720)
= 26.87 V

So, the correct answer for part b) is approximately 26.87 V.