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Capacitance is a fundamental property of capacitors. It defines a capacitor’s ability to store an electric charge. When you think of capacitance, imagine a container filling up with liquid. The bigger the container, the more it can hold. Capacitors work similarly but with electric charge. The unit of capacitance is the farad (F). However, due to the large size of one farad, practical capacitors often have values in microfarads (\( \mu F \)) or picofarads.
Capacitors are characterized by two main things:

• Plate Size: Larger plates can store more charge.
• Distance Between Plates: A smaller distance increases capacitance.

Capacitance is determined using the formula \( C = \frac{q}{V} \), where \( q \) is the charge stored, and \( V \) is the voltage applied. Understanding this relation helps clarify why capacitor B, with its given capacitance of \(6.7 \mu F\), stores a specific amount of charge when a certain voltage is applied. The higher the capacitance, for a given voltage, the more charge a capacitor can store. This is because a high capacitance allows more electric field lines to be stored between the capacitor plates.

Charge storage in capacitors is an essential concept in electrostatics. Capacitors store energy as electric charge, ready to be discharged when needed, similar to a battery. The stored charge in a capacitor is directly proportional to the voltage applied across it. This phenomenon is described by the equation \( q = C \times V \), where \( q \) is the charge, \( C \) is the capacitance, and \( V \) is the voltage.

• Higher voltage or higher capacitance will result in more charge being stored.
• The amount of charge doesn't depend solely on the capacitance or voltage but on their product.

In this specific exercise, capacitor B stores a charge calculated as \(q_B = C_B \times V_B\). This charge storage capacity is why capacitors are so valuable in circuits, as they allow for the rapid storage and discharge of energy. Understanding how the capacitance and voltage influence charge storage can help predict the behavior of capacitor B in this scenario.

The energy stored in capacitors is one of their most useful properties. When a charge is stored in a capacitor, energy is also stored. This energy can be used to power devices momentarily when required. The relationship between energy stored in a capacitor, the charge, and the voltage is given by the equation: \( E = \frac{1}{2} C V^2 \).

• Energy increases with greater charge and greater voltage.
• The energy is a quadratic function of voltage, meaning that doubling the voltage quadruples the stored energy.

Capacitor A in this exercise stores an energy of \(5.0 \times 10^{-5}\, \mathrm{J}\) thanks to the charge \(11 \mu C\) and its capacitance. It demonstrates how changes in voltage affect energy storage. Calculating the energy helps engineers and scientists anticipate how much power a capacitor can deliver. For capacitor B, although we didn't directly calculate its energy, understanding its capacitance and the applied voltage helps assess the potential energy it can store.