Question

An air-filled parallel-plate capacitor with plate separation \(d\) and plate area \(A\) is connected to a battery that applies a voltage \(V_{0}\) between plates. With the battery left connected, the plates are moved apart to a distance of \(10 d\). Determine by what factor each of the following quantities changes: \((a) V_{0} ;(b) C ;(c) E ;(d) D ;(e) Q ;(f) \rho_{S} ;(g) W_{E}\)

Short answer

When the plate separation of an air-filled parallel-plate capacitor connected to a constant-voltage battery increases from \(d\) to \(10d\), the following factors change: (a) Voltage (\(V_0\)): Remains unchanged (factor of 1). (b) Capacitance (\(C\)): Decreases by a factor of 10. (c) Electric field (\(E\)): Decreases by a factor of 10. (d) Electric displacement field (\(D\)): Decreases by a factor of 10. (e) Charge (\(Q\)): Decreases by a factor of 10. (f) Surface charge density (\(\rho_S\)): Decreases by a factor of 10. (g) Energy stored in the capacitor (\(W_E\)): Decreases by a factor of 10.

Step-by-step solution

Voltage (\(V_0\))

The battery maintains a constant voltage (\(V_0\)) between the plates. So, when the plates are moved apart, the voltage between the plates does not change because the battery is still connected. Therefore, the factor by which the voltage changes is 1.

Capacitance (\(C\))

The capacitance formula for a parallel-plate capacitor is \(C = \frac{\epsilon_0 A}{d}\). When the plate separation is increased to \(10d\), the new capacitance \(C'\) will be: $$C' = \frac{\epsilon_0 A}{10d}.$$ Now, we can find the factor by which capacitance changes by dividing \(C'\) by \(C\): $$\frac{C'}{C} = \frac{\frac{\epsilon_0 A}{10d}}{\frac{\epsilon_0 A}{d}} = \frac{1}{10}.$$ The capacitance decreases by a factor of 10.

Electric field (\(E\))

The electric field between the plates of a charged capacitor is given by \(E=\frac{V}{d}\), where \(V\) is the voltage across the plates. Since the voltage remains constant and the plate separation increases to \(10d\), the new electric field \(E'\) will be: $$E'=\frac{V_0}{10d}.$$ To find the factor by which the electric field changes, we can divide \(E'\) by \(E\): $$\frac{E'}{E}=\frac{\frac{V_0}{10d}}{\frac{V_0}{d}}=\frac{1}{10}.$$ The electric field decreases by a factor of 10.

Electric displacement field (\(D\))

The electric displacement field between the plates of a capacitor in a vacuum is given by \(D=\epsilon_0 E\), where \(\epsilon_0\) is the vacuum permittivity and \(E\) is the electric field. Since the electric field decreases by a factor of 10, the electric displacement field will also decrease by the same factor. Thus, the factor by which the electric displacement field changes is \(\frac{1}{10}\).

Charge (\(Q\))

The charge on the capacitor is given by \(Q=CV\) for a capacitor with a constant voltage. Since the capacitance decreases by a factor of 10, and the voltage remains constant, the charge will also decrease by a factor of 10. Thus, the factor by which the charge changes is \(\frac{1}{10}\).

Surface charge density (\(\rho_S\))

The surface charge density on the plates of a capacitor is given by \(\rho_S=\frac{Q}{A}\), where \(Q\) is the charge and \(A\) is the area of the plates. Since the charge decreases by a factor of 10 and the area remains constant, the surface charge density will also decrease by a factor of 10. Thus, the factor by which the surface charge density changes is \(\frac{1}{10}\).

Energy stored in the capacitor (\(W_E\))

The energy stored in the capacitor is given by \(W_E=\frac{1}{2}CV^2\). Since the capacitance decreases by a factor of 10 and the voltage remains constant, the energy stored will also decrease by a factor of 10. Thus, the factor by which the energy stored changes is \(\frac{1}{10}\). In summary: - The voltage \(V_0\) remains unchanged (factor of 1). - The capacitance \(C\) decreases by a factor of 10. - The electric field \(E\) decreases by a factor of 10. - The electric displacement field \(D\) decreases by a factor of 10. - The charge \(Q\) decreases by a factor of 10. - The surface charge density \(\rho_S\) decreases by a factor of 10. - The energy stored in the capacitor \(W_E\) decreases by a factor of 10.

Key concepts

parallel-plate capacitor

A parallel-plate capacitor consists of two conductive plates placed parallel to each other, with a certain distance separating them. This setup is designed to store electrical energy efficiently. Capacitors can be made up of various materials to achieve different properties, but in this exercise, we are focusing on an air-filled parallel-plate capacitor.

When a voltage is applied across the plates, an electric field is established between them. This electric field causes the capacitor to store energy. The separation between the plates and their area are critical factors. They significantly influence the capacitance, which is a measure of the capacitor's ability to store charge for a given voltage.

• The closer the plates, the higher the capacitance.
• The larger the plate area, the higher the capacitance.

The operation of a parallel-plate capacitor is straightforward: it can quickly store and release energy, making it an essential component in many electronic circuits.

electric field

The electric field in a parallel-plate capacitor is the force field created between the plates when the capacitor is charged. It is uniform in the space between the plates, assuming that the plates are large and closely spaced compared to their separation distance.

The strength of the electric field \( E \) is directly related to the voltage \( V \) applied across the plates and the distance \( d \) separating them, given by the formula:
\[ E = \frac{V}{d} \]
This means that the electric field strength is influenced by how much voltage is applied and how far apart the plates are. For instance, in this exercise, when the plates are moved further apart (increasing \( d \)), the electric field decreases because the same voltage now influences a larger distance. Consequently, the electric field decreases by a factor of 10 when the distance is increased to \( 10d \).

Maintaining the uniformity of the electric field between the plates ensures the efficiency of energy storage in a capacitor.

capacitance

Capacitance is a measure of how much electric charge a capacitor can store at a given voltage. It's a fundamental property of capacitors that determines their ability to hold energy.

For a parallel-plate capacitor, the capacitance \( C \) is determined by the formula:
\[ C = \frac{\epsilon_0 A}{d} \]
where:

• \( \epsilon_0 \) is the permittivity of free space,
• \( A \) is the plate area,
• \( d \) is the separation between the plates.

This formula shows that increasing the plate area \( A \) or decreasing the distance \( d \) increases the capacitance. Conversely, if the separation \( d \) is increased, as in the exercise where it is set to \( 10d \), the capacitance decreases by a factor of 10. Here, the permittivity \( \epsilon_0 \) remains constant because the dielectric is air.

Understanding capacitance is crucial, especially when designing circuits that require precise energy storage and release.

electric displacement field

The electric displacement field, often represented as \( D \), is an important concept that describes how electric fields interact with materials. In vacuum or air-filled capacitors, it is closely related to the electric field \( E \).

The relationship is given by the equation:
\[ D = \epsilon_0 E \]
where \( \epsilon_0 \) is the permittivity of free space. This relation states that \( D \) is proportional to \( E \), meaning that changes in the electric field directly affect the electric displacement field.

In our scenario, as the plates' separation increases, both the electric field \( E \) and electric displacement field \( D \) decrease by a factor of 10. Understanding \( D \) is essential, as it provides insights into how a material responds to an electric field and helps confirm calculations related to surface charge and field strengths in capacitors.

surface charge density

Surface charge density, denoted as \( \rho_S \), describes how charge is distributed across the surface of the capacitor's plates. It is a measure of the amount of charge per unit area and is key in determining the behavior of capacitors.

The formula for surface charge density is:
\[ \rho_S = \frac{Q}{A} \]
where \( Q \) is the total charge on the plates, and \( A \) is the area of the plates. This ratio tells us about how concentrated or spread out the charge is on the surface.

In the context of our exercise, as the capacitance decreases by a factor of 10 due to increased separation between the plates, the stored charge \( Q \) also decreases by the same factor because the voltage remains constant. Consequently, the surface charge density \( \rho_S \) decreases by a factor of 10, as the area stays the same. These changes highlight how shifts in physical parameters, like plate separation, directly impact the charge distribution.