A 170 -pF capacitor is connected in series with an unknown capacitance, and as a series combination they are connected to a battery with an emf of 25.0 V. If the 170 -pF capacitor stores 135 pC of charge on its plates, what is the unknown capacitance?

Q1 = C1V1.

V1 = Q1 / C1. = 135 / 170 = 0.794 Volts.

V2 = 25-0.794 = 24.206 Volts.
Q2 = Q1 = 135pC.
Q2 = C2V2.
C2 = Q2/V2 = 135 / 24.206 = 5.577 pF.

To find the unknown capacitance, we can use the formula for capacitance in series connection:

1/Ceq = 1/C1 + 1/C2

Where C1 and C2 are the capacitances of the known and unknown capacitors, and Ceq is the equivalent capacitance of the series combination.

We are given that the known capacitance C1 is 170 pF and the charge Q1 on its plates is 135 pC. We need to convert these values to Farads:

C1 = 170 pF = 170 × 10^(-12) F
Q1 = 135 pC = 135 × 10^(-12) C

Now, we can find the equivalent capacitance Ceq using the formula:

Ceq = Q1 / V

Where V is the voltage across the series combination. We are given that the battery has an emf of 25.0 V.

Ceq = 135 × 10^(-12) C / 25.0 V

Simplify the expression:

Ceq = 5.4 × 10^(-12) F

Now, plug in the values into the formula for the series connection:

1/Ceq = 1/C1 + 1/C2

1/(5.4 × 10^(-12)) = 1/(170 × 10^(-12)) + 1/C2

Solve the equation to find C2:

1/C2 = 1/(5.4 × 10^(-12)) - 1/(170 × 10^(-12))

1/C2 = (170 - 5.4) / (170 × 5.4 × 10^(-12))

C2 = 1 / [(170 - 5.4) / (170 × 5.4 × 10^(-12))]

C2 ≈ 31.61 pF

Therefore, the unknown capacitance is approximately 31.61 pF.