Question

The space between the plates of an isolated parallel plate capacitor is filled with a slab of dielectric material. The magnitude of the charge \(Q\) on each plate is kept constant. If the dielectric material is removed from between the plates, the energy stored in the capacitor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

Short answer

Answer: a) increases

Step-by-step solution

1. Recall the formula for the energy stored in a capacitor

The energy \(U\) stored in a capacitor can be calculated using the formula \(U = \dfrac{1}{2} QV\), where \(Q\) is the charge on each plate and \(V\) is the voltage across the plates.

2. Relate the voltage to the capacitance and charge

We can use the formula \(C = \dfrac{Q}{V}\) to express the voltage as \(V = \dfrac{Q}{C}\).

3. Express the energy in terms of capacitance and charge

Substituting the expression for voltage from step 2 into the energy formula, we get: \(U = \dfrac{1}{2} Q\left(\dfrac{Q}{C}\right) = \dfrac{Q^2}{2C}\).

4. Find the difference in capacitance with and without the dielectric

If a dielectric material is inserted between the plates of a capacitor, the capacitance is increased by a factor called the dielectric constant, denoted by \(k\). Therefore, the capacitance with the dielectric is \(C_k = kC_0\), where \(C_0\) is the capacitance without the dielectric.

5. Calculate the difference in the energy stored in the capacitor with and without the dielectric

The energy stored in the capacitor with the dielectric is \(U_k = \dfrac{Q^2}{2C_k} = \dfrac{Q^2}{2kC_0}\), whereas, without the dielectric, it is \(U_0 = \dfrac{Q^2}{2C_0}\). Now, let's divide both energies: \(\dfrac{U_0}{U_k} = \dfrac{\dfrac{Q^2}{2C_0}}{\dfrac{Q^2}{2kC_0}} = k\). Since \(k>1\), we can conclude that \(U_0 > U_k\).

6. Determine the change in the stored energy when the dielectric is removed

Since \(U_0 > U_k\), the energy stored in the capacitor increases when the dielectric material is removed. So, the correct answer is: a) increases.

Key concepts

Dielectric Material

Dielectric materials are crucial in the functioning of capacitors. They are insulating substances placed between the plates of a capacitor, serving multiple roles.
The primary role is to increase a capacitor's capacitance.

• By inserting a dielectric, the capacitance increases by a factor known as the dielectric constant, indicated symbolically as \(k\).
• This factor is always greater than 1, meaning that dielectrics enhance the efficiency of capacitors.

Using dielectrics enables capacitors to store more electrical energy within the same physical space.
This is incredibly useful in various electronic devices, making them more compact and powerful.

Energy Storage

In capacitors, energy storage is related to both the charge on the plates and the voltage across them. The energy \(U\), stored in a capacitor, is expressed as \(U = \frac{1}{2} QV\).
This equation shows that the energy stored increases with either an increase in charge \(Q\) or voltage \(V\). Understanding energy storage is key in applications like temporary power supply in circuits.

• Capacitors can discharge quickly, providing bursts of energy when needed.
• Ensuring efficient storage and release of energy enhances the performance of electronic systems.

Capacitance Change

The capacitance of a capacitor changes based on the presence of dielectric materials. Using the formula \(C = \frac{Q}{V}\), we identify how capacitance \(C\) is related to charge and voltage.

• Capacitance increases with a dielectric, represented mathematically as \(C_k = kC_0\), where \(C_0\) is the original capacitance without the dielectric.
• When the dielectric is removed, capacitance decreases, resulting in a different energy storage capacity.

This dynamic allows engineers to control and modify how capacitors perform in various circuits under different conditions.
Being able to predict changes in capacitance is critical for designing efficient electronic components.

Voltage Relationship

When considering capacitors, voltage plays a crucial role in their operation. The relationship between voltage and capacitance is inversely proportional when charge is kept constant.
By using the formula \(V = \frac{Q}{C}\), we see:

• Increasing capacitance with a dielectric reduces voltage if charge does not change.
• Lifting the dielectric increases voltage since capacitance decreases, given the formula \(V = \frac{Q}{C_0}\).

Understanding these relationships helps in predicting how a capacitor will behave when dielectric materials are introduced or removed, impacting the overall energy stored and the voltage across the plates.