Problem 21.40


A pair of 10uF capacitors in a high-power laser are charged to 1.7kV.


Paige Nieuwenhuis

Part A -

What charge is stored in each capacitor?

Express your answer using two significant figures.




Part B -

How much energy is stored in each capacitor?

Express your answer using two significant figures.


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Try Again; 4 attempts remaining



Part B -

If the plates are separated by a Teflon sheet, how thick is the sheet?

Express your answer using two significant figures.

A. Use the equation Q = C*V

You don't say if the capacitors are in parallel or series. If in parallel, V = 1700 V for each.

B. E = (1/2) C V^2

Your second "part B" must belong to a different question. More facts are needed about capacitor size, capacitance, and whether the Teflon fills the gap.

To solve this problem, we need to use the following formulas:

1) Q = CV - This formula relates the charge (Q) stored in a capacitor to its capacitance (C) and the voltage (V) across it.

2) Energy stored in a capacitor (E) = 1/2 * CV^2 - This formula relates the energy stored in a capacitor to its capacitance and the voltage across it.

Let's solve Part A first:

Given:
Capacitance, C = 10 uF = 10 * 10^-6 F (convert units to Farads)
Voltage, V = 1.7 kV = 1.7 * 10^3 V (convert units to Volts)

Using the formula Q = CV, we can calculate the charge stored in each capacitor:

Q = (10 * 10^-6 F) * (1.7 * 10^3 V) = 17 * 10^-3 C = 0.017 C

Therefore, the charge stored in each capacitor is 0.017 C.

Now let's solve Part B:

Using the formula for energy stored in a capacitor, E = 1/2 * CV^2, we can calculate the energy stored in each capacitor:

E = 1/2 * (10 * 10^-6 F) * (1.7 * 10^3 V)^2 = 1/2 * 10 * 10^-6 * (1.7 * 1.7 * 10^3 * 10^3) J
E = 1/2 * 10 * 10^-6 * 2.89 * 10^6 J = 14.45 J

Therefore, the energy stored in each capacitor is 14.45 J.

Now let's move on to Part C:

Unfortunately, the information given is not sufficient to calculate the thickness of the Teflon sheet. We need more information such as the dielectric constant or the capacitance value with the Teflon sheet present. Without this additional information, it is not possible to determine the thickness of the Teflon sheet.

To solve Part A, we can use the formula for the charge stored in a capacitor:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage.

Given that the capacitance of each capacitor is 10uF (microfarads) and the voltage is 1.7kV (kilovolts), we can substitute these values into the formula:

Q = (10uF)(1.7kV)

Now, let's convert the units:

10uF = 10 * 10^-6 F (microfarads to farads)

1.7kV = 1.7 * 10^3 V (kilovolts to volts)

Q = (10 * 10^-6 F)(1.7 * 10^3 V)

Now, calculate the product of these values:

Q = 17 * 10^-3 C

Rounding to two significant figures, the charge stored in each capacitor is approximately 0.017 C.

To solve Part B, we can use the formula for the energy stored in a capacitor:

E = 0.5 * CV^2

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

Using the same values for capacitance and voltage as in Part A, we can substitute them into the formula:

E = 0.5 * (10uF)(1.7kV)^2

Again, let's convert the units:

10uF = 10 * 10^-6 F (microfarads to farads)

1.7kV = 1.7 * 10^3 V (kilovolts to volts)

E = 0.5 * (10 * 10^-6 F)(1.7 * 10^3 V)^2

Now, calculate the square of the voltage and multiply it with the capacitance and the constant factor:

E = 0.5 * (10 * 10^-6 F)(1.7 * 10^3 V)^2

E = 0.5 * (10 * 10^-6 F)(2.89 * 10^6 V^2)

E = 14.45 J

Rounding to two significant figures, the energy stored in each capacitor is approximately 14 J.

For Part C, the thickness of the Teflon sheet cannot be determined with the given information.