91Ó°ÊÓ

When a capacitor stores energy, it does so by holding an electric charge on its plates. The energy stored in a capacitor can be calculated using the formula: \[ U = \frac{1}{2} C V^2 \] where:

• \( U \) is the energy stored in joules (J),
• \( C \) represents the capacitance in farads (F), and
• \( V \) is the potential difference across the capacitor in volts (V).

In the given problem, before introducing a dielectric material, the capacitor had a capacitance of \(12.5 \mu F\) and a potential difference of \(24.0\, \text{V}\). The initial energy stored can be computed by substituting these values into the formula.
After inserting the dielectric, the capacitance changes, affecting the energy stored. With a dielectric, the capacitor can hold more energy, visible from the increase in stored energy calculated after the dielectric insertion. This change is determined by the dielectric constant, which allows the capacitor to store more energy even though the potential difference remains the same.

The dielectric constant, denoted as \( \kappa \), is a crucial factor in understanding how materials affect capacitors. It represents how much more capacitance can be achieved by filling a capacitor completely with a dielectric material compared to a vacuum.

• A higher dielectric constant means a material can store more charge at the same voltage.
• This constant is dimensionless, providing a ratio that compares the insulating ability of the dielectric to the capacitance in a vacuum.
• Common dielectric constants for materials can vary, helping select the right material for specific applications.

In the exercise, the dielectric constant is \(3.75\), indicating that the capacitance becomes 3.75 times its initial value when the dielectric is inserted. This directly affects the energy stored because the increased capacitance leads to more energy stored in the capacitor.

The potential difference, also known as voltage, across a capacitor is the driving force that causes charged particles to accumulate on its plates.

• It is measured in volts (V), representing the work needed to move a unit charge between the plates.
• In the context of a capacitor, the potential difference remains constant even when a dielectric is introduced.
• This constancy ensures that the electric field between the plates remains the same, focusing on how a dielectric affects capacitance and stored energy.

In the given exercise, the potential difference across the capacitor plates is consistently \(24.0\, \text{V}\). Maintaining this voltage is vital because when it's applied across a capacitor with a temporarily varying capacitance, like when adding a dielectric, it determines how the stored energy changes. The voltage doesn't change, but the increased capacitance boosts the energy storage capacity.