how does the total energy stored in a capacitor change, when the medium of air is replaced by a medium of dielectric constaant K? Explain.
Same charge, not hooked to an outside source?
Energy=1/2 Q^2/Ck where C is the dry air capacitance, and k is the dielectric costant. Hmmm,so energy stored is reduced. That must mean it did work on the person trying to force the dielectric into the gap.
What if it is constant voltage, hooked to a voltage source?
Energy=1/2 CkV^2 now energy is increased, which means the outside circuit must have suppied energy.
To understand how the total energy stored in a capacitor changes when the medium of air is replaced by a medium of dielectric constant K, let's break down the explanation into a few steps:
Step 1: Understanding the energy stored in a capacitor in air
In a capacitor, the energy is stored in the electric field between the plates. When a capacitor is charged, the electric field is established between the positive and negative charges on the plates. The energy stored in the capacitor is given by the formula:
E = (1/2) * C * V^2
Where:
E is the energy stored in the capacitor
C is the capacitance of the capacitor
V is the voltage across the capacitor
Step 2: How the introduction of a dielectric affects the electric field
When a dielectric material is introduced between the plates of a capacitor, it has the ability to polarize, meaning it aligns its charges in response to an external electric field. This polarization leads to the formation of an opposing electric field within the dielectric, thus reducing the effective electric field between the plates.
Step 3: Effect of the dielectric on capacitance
The introduction of a dielectric material increases the capacitance of the capacitor. Capacitance with a dielectric material can be given by:
C' = K * C
Where:
C' is the capacitance with the dielectric material
K is the dielectric constant of the material
C is the capacitance without the dielectric material (in air)
Step 4: Calculating the energy stored with the dielectric material
Using the formula for energy stored in a capacitor, we can substitute the capacitance with the dielectric material into the equation. Since the capacitance increases with the dielectric constant K, the energy stored in a capacitor with a dielectric material becomes:
E' = (1/2) * K * C * V^2
Where:
E' is the energy stored in the capacitor with the dielectric material
C is the original capacitance (without the dielectric material)
V is the voltage across the capacitor
Step 5: Comparing the energy stored in air versus with a dielectric
By comparing the two energy equations, we can see that the energy stored in the capacitor increases when a dielectric material with a higher dielectric constant is introduced. The energy is directly proportional to the dielectric constant K. Therefore, replacing air with a medium of dielectric constant K leads to an increase in the total energy stored in the capacitor.
In summary, when the medium of air is replaced by a medium with a dielectric constant K, the energy stored in the capacitor increases due to the increased capacitance caused by the dielectric material.