Question:

A parallel plate capacitor is connected to a battery of 12V, causing the plates to become charged. The separation between the plates is 0.02m and the area of each plate is 0.005m². The capacitor is then disconnected from the battery and connected to another identical capacitor in series. Determine:

(a) The capacitance of the individual capacitors in parallel.

(b) The total capacitance when the capacitors are connected in series.

(c) The charge stored on each capacitor when connected in parallel.

(d) The potential difference across each capacitor when connected in series.

(e) The energy stored in each capacitor when connected in parallel.

(f) The energy stored in each capacitor when connected in series.

Answer:

(a) The capacitance of the individual capacitor in parallel can be determined using the formula:

$C = \frac{Q}{V}$

where: $C$ = capacitance (in Farads), $Q$ = charge stored on the capacitor (in Coulombs), $V$ = potential difference across the capacitor (in Volts).

We know that the voltage across the capacitor is 12V and the charge stored on it is given by $Q = CV$. Substituting the given values, we have:

$C = \frac{Q}{V} = \frac{CV}{V}$

$C = \frac{12V}{12V} = 1 \, \text{Farad}$

Therefore, the capacitance of each individual capacitor in parallel is 1 Farad.

(b) When the capacitors are connected in series, the total capacitance can be determined using the formula:

$\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2}$

Since the capacitors are identical, we can rewrite the equation as:

$\frac{1}{C_{total}} = \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_{total}} = \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_{total}} = \frac{2}{C}$

$C_{total} = \frac{C}{2}$

Substituting the known value of $C$ (1 Farad), we find:

$C_{total} = \frac{1 \, \text{F}}{2} = 0.5 \, \text{Farads}$

Therefore, the total capacitance when the capacitors are connected in series is 0.5 Farads.

(c) When the capacitors are connected in parallel, the charge stored on each capacitor is the same. Since the total charge stored on the two capacitors is the same as that of the original capacitor, we can use the equation $Q = CV$ to find the charge stored on each capacitor.

$Q_{parallel} = CV_{parallel} = (1 \, \text{F})(12V) = 12 \, \text{Coulombs}$

Therefore, each capacitor stores 12 Coulombs of charge when connected in parallel.

(d) When the capacitors are connected in series, the potential difference across each capacitor is the same. Since the total potential difference across the two capacitors is equal to the original potential difference (12V), we can divide the voltage equally between the two capacitors.

$V_{series} = \frac{V_{original}}{2} = \frac{12V}{2} = 6V$

Therefore, the potential difference across each capacitor when connected in series is 6 Volts.

(e) The energy stored in a capacitor can be determined using the formula:

$U = \frac{1}{2} CV^2$

When the capacitors are connected in parallel, the total energy stored can be found by adding the energy of each individual capacitor.

$U_{parallel} = \frac{1}{2} C_{total} V_{parallel}^2$

Substituting the known values, we have:

$U_{parallel} = \frac{1}{2} (1F)(12V)^2 = 72 \, \text{Joules}$

Therefore, each capacitor stores 72 Joules of energy when connected in parallel.

(f) When the capacitors are connected in series, the energy stored in each capacitor can be determined using the same formula:

$U_{series} = \frac{1}{2} C_{total} V_{series}^2$

Substituting the known values, we get:

$U_{series} = \frac{1}{2} (0.5F)(6V)^2 = 9 \, \text{Joules}$

Therefore, each capacitor stores 9 Joules of energy when connected in series.