Question

Two parallel plate capacitors, \(C_{1}\) and \(C_{2}\), are connected in series to a 96.0-V battery. Both capacitors have plates with an area of \(1.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm} ; C_{1}\) has air between its plates, and \(C_{2}\) has that space filled with porcelain (dielectric constant of 7.00 and dielectric strength of \(5.70 \mathrm{kV} / \mathrm{mm})\) a) After charging, what are the charges on each capacitor? b) What is the total energy stored in the two capacitors? c) What is the electric field between the plates of \(C_{2} ?\)

Short answer

Answer: When two parallel plate capacitors are connected in series, the charge distribution on each capacitor is the same as the charge on the equivalent capacitor, i.e., \(Q_{1} = Q_{2} = Q_{eq}\).

Step-by-step solution

Determine the equivalent capacitance for the series connection

To determine the equivalent capacitance (\(C_{eq}\)) for two capacitors connected in series, we use the formula: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\)

Calculate the capacitance for each capacitor

To find the capacitance of each capacitor, we need to use the formula for the capacitance of a parallel plate capacitor: \(C = \frac{\varepsilon_{0} \varepsilon A}{d}\) where \(\varepsilon_{0} = 8.85 \cdot 10^{-12} \mathrm{F/m}\) is the vacuum permittivity, \(\varepsilon\) is the relative permittivity (dielectric constant), \(A\) is the plate area, and \(d\) is the plate separation. For \(C_{1}\) (air-filled), we'll use \(\varepsilon = 1\) whereas, for \(C_{2}\) (porcelain-filled), we'll use the given dielectric constant of 7.

Calculate the equivalent capacitance

Using the calculated values of \(C_1\) and \(C_2\), calculate the equivalent capacitance \(C_{eq}\).

Calculate the charge on the equivalent capacitor

Now, use the equivalent capacitance to calculate the charge on the equivalent capacitor using the formula: \(Q_{eq} = C_{eq}V\) where \(V = 96.0 \mathrm{V}\) is the battery voltage.

Determine the charge on each capacitor

Since the capacitors are connected in series, the charge on each capacitor will be the same as on the equivalent capacitor: \(Q_{1} = Q_{2} = Q_{eq}\)

Calculate the total energy stored in the capacitors

The total stored energy can be found using the formula for the energy stored in a capacitor: \(U = \frac{1}{2}CV^2\) We can find the individual energy stored in each capacitor and then add them together to get the total energy. Alternatively, we can use equivalent capacitance and total voltage: \(U_{total} = \frac{1}{2}C_{eq}V_{total}^2\)

Calculate the electric field between the plates of \(C_{2}\)

To find the electric field between the plates of \(C_2\), we'll use the formula: \(E = \frac{Q}{\varepsilon_{0} \varepsilon A}\) where \(Q\) is the charge on the capacitor, \(\varepsilon_{0}\) is the vacuum permittivity, \(\varepsilon\) is the relative permittivity (dielectric constant), and \(A\) is the plate area. Use the given values and the calculated charge on \(C_2\) to find the electric field.