Two identical capacitors are hooked in series to a 16.6 V emf. If the charge stored on any one of the capacitors is 0.88 micro-Coulombs after the capacitors are fully charged, what is the capacitance of either one of the capacitors?

The charge on each is the same.

The voltage on each is half the battery = 8.3 volts

C = q/V

C = .88 *10^-6 /8.3

To find the capacitance of either one of the capacitors, we can make use of the formula for the equivalent capacitance of capacitors connected in series, which is given as:

1/C_eq = 1/C_1 + 1/C_2 + ...

Since the two capacitors in this case are identical, the equation simplifies to:

1/C_eq = 1/C_1 + 1/C_1

Let's denote the capacitance of one of the capacitors as C. Plugging this value into the equation, we get:

1/C_eq = 1/C + 1/C

Simplifying further:

1/C_eq = 2/C

Next, we rearrange the equation to solve for C_eq:

C_eq = C/2

Now, we are given the charge stored on any one of the capacitors, which we'll denote as Q. The charge stored on each capacitor can be calculated using the formula:

Q = C_eq * emf

Plugging in the values we have:

0.88 micro-Coulombs = (C/2) * 16.6 V

To solve for C, we isolate the variable:

C = (0.88 micro-Coulombs * 2) / 16.6 V

Calculating this expression, we find:

C ≈ 0.053 micro-Farads

Therefore, the capacitance of either one of the capacitors is approximately 0.053 micro-Farads.