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A capacitor is a device that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material called a dielectric.
When a voltage is applied across the plates, an electric charge builds up on each plate, with one plate becoming positively charged and the other negatively charged.
Capacitors are widely used in electronic circuits to store energy, filter signals, and stabilize voltage and power flow.
The ability of a capacitor to store charge is measured by its capacitance, which is defined as the amount of charge stored per unit voltage applied. The unit of capacitance is the farad (F).

The potential difference, also known as voltage, is the difference in electric potential between two points in a circuit.
It is the driving force that pushes electric charges through a conductor. Voltage is measured in volts (V).
When a capacitor is connected to a battery or power source, the potential difference causes an accumulation of charge on the capacitor's plates.
In our exercise, the potential difference between the plates of the first capacitor is initially 50 V, which drops to 35 V after connecting it in parallel with a second capacitor.
The change in potential difference is crucial for determining the new charge distribution and calculating the capacitance of the second capacitor.

Calculating the charge on a capacitor involves using the formula: Or in LaTeX: \( Q = C \times V \)
Where:

• \( Q \) is the charge in coulombs (C).
• \( C \) is the capacitance in farads (F).
• \( V \) is the potential difference in volts (V).

In our exercise, we first need to find the initial charge \( Q \) on the first capacitor using \( C_1 = 100 \, \mathrm{pF} \) and \( V_1 = 50 \, \mathrm{V} \),
Which gives: \[ Q_1 = C_1 \times V_1 = 100 \, \mathrm{pF} \times 50 \, \mathrm{V} = 5000 \, \mathrm{pC} \]
After connecting in parallel with the second uncharged capacitor, the potential difference drops to 35 V.
Calculate the new charge \( Q_f \) on the first capacitor: \[ Q_f = C_1 \times V_f = 100 \, \mathrm{pF} \times 35 \, \mathrm{V} = 3500 \, \mathrm{pC} \]
Subtract \( Q_f \) from \( Q_1 \) to find the charge that has moved to the second capacitor: \[ Q_{C2} = Q_1 - Q_f = 5000 \, \mathrm{pC} - 3500 \, \mathrm{pC} = 1500 \, \mathrm{pC} \]

When capacitors are connected in parallel, the total capacitance \( C_{total} \) is the sum of the individual capacitances:
Or in LaTeX: \( C_{total} = C_1 + C_2 \)
In a parallel connection, each capacitor experiences the same potential difference.
In our exercise, after the first charged capacitor (100 pF, 50 V) is connected in parallel with the second initially uncharged capacitor, the potential difference equalizes at 35 V.
To find the capacitance of the second capacitor \( C_2 \),
we use the fact that the charge moving to the second capacitor is 1500 pC and the new potential difference is 35 V:
Solve \( Q = C \times V \) for \( C_2 \): \[ C_2 = \frac{Q_{C2}}{V_f} = \frac{1500 \, \mathrm{pC}}{35 \, \mathrm{V}} \approx 42.86 \, \mathrm{pF} \]
So, the capacitance of the second capacitor is approximately 42.86 pF.
This example demonstrates how connecting capacitors in parallel allows them to share charge and equalize voltage, providing useful functionality in various electrical circuits.