Question

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is \(32.0 \mathrm{pF}\) when the separation between plates is \(d=0.500 \mathrm{~cm} .\) a) A battery with potential difference \(V=9.0 \mathrm{~V}\) is connected to the plates. What is the charge distribution, \(\sigma\), on the left plate? What are the capacitance, \(C^{\prime},\) and charge distribution, \(\sigma^{\prime},\) when \(d\) is changed to \(0.250 \mathrm{~cm} ?\) b) With \(d=0.500 \mathrm{~cm}\), the battery is disconnected from the plates. The plates are then moved so that \(d=0.250 \mathrm{~cm}\) What is the potential difference \(V^{\prime},\) between the plates?

Short answer

**a) Calculating charge distribution, \(\sigma\)** First, we need to find the charge, Q, on the capacitor when the potential difference is 9.0 V, and the capacitance is 32.0 pF. Calculate the charge: \(Q = (32.0 \times 10^{-12}) \times 9.0\) \(Q = 288 \times 10^{-12} C\) **Find the area, A, of the plates** Calculate the area: \(A = (32.0 \times 10^{-12}) \times \frac{(0.5\times 10^{-2})}{(8.85 \times 10^{-12})}\) \(A = 1.8 \times 10^{-4} m^2\) **Calculate the charge distribution, \(\sigma\)** By plugging in the values for Q and A, we will find the charge distribution on the left plate: \(\sigma = \frac{(288 \times 10^{-12})}{(1.8 \times 10^{-4})}\) \(\sigma = 1.6 \times 10^{-6} C/m^2\) **Calculate the new capacitance, \(C^\prime\)** Calculate the new capacitance, \(C^\prime\): \(C^\prime = \varepsilon_0 \frac{(1.8 \times 10^{-4})}{(0.25 \times 10^{-2})}\) \(C^\prime = 64.0 \times 10^{-12} F\) **Calculate the new charge distribution, \(\sigma^\prime\)** Use the new capacitance \(C^\prime\) value to calculate the new charge distribution, \(\sigma^\prime\): \(\sigma^\prime = \frac{(288 \times 10^{-12})}{(1.8 \times 10^{-4})}\) \(\sigma^\prime = 1.6 \times 10^{-6} C/m^2\) **b) Determine the potential difference \(V^\prime\)** Plug in the values of Q and \(C^\prime\) to calculate the potential difference, \(V^\prime\): \(V^\prime = \frac{(288 \times 10^{-12})}{(64.0 \times 10^{-12})}\) \(V^\prime = 4.5 V\) Thus, the charge distribution on the plates remains the same at \(1.6 \times 10^{-6} C/m^2\) when the separation is changed, and the potential difference decreases from 9.0 V to 4.5 V when the battery is disconnected and the plates are moved closer.

Step-by-step solution

a) Calculating charge distribution, \(\sigma\)

First, we need to find the charge, Q, on the capacitor when the potential difference is 9.0 V, and the capacitance is 32.0 pF. We will use the formula: \(Q = C \times V\) Converting 32.0 pF to F: \(C = 32.0 \times 10^{-12} F\) Now, calculate the charge: \(Q = (32.0 \times 10^{-12}) \times 9.0\)

Find the area, A, of the plates

To find the charge distribution, we need the area of the plates. We will use the capacitance formula: \(C = \varepsilon_0 \frac{A}{d}\) Rearranging the equation, we have: \(A = C \times \frac{d}{\varepsilon_0}\) Now, plug in the values: \(d = 0.5 \times 10^{-2} m\) \(\varepsilon_0 = 8.85 \times 10^{-12} F/m\) Calculate the area: \(A = (32.0 \times 10^{-12}) \times \frac{(0.5\times 10^{-2})}{(8.85 \times 10^{-12})}\)

Calculate the charge distribution, \(\sigma\)

Now we can find the charge distribution using the formula: \(\sigma = \frac{Q}{A}\) By plugging in the values for Q and A, we will find the charge distribution on the left plate.

Calculate the new capacitance, \(C^\prime\)

Now let's find the new capacitance when the separation between the plates is changed to 0.250 cm. Again, using the capacitance formula, \(C^\prime = \varepsilon_0 \frac{A}{d^\prime}\) With the new value for d: \(d^\prime = 0.25 \times 10^{-2} m\) Calculate the new capacitance, \(C^\prime\).

Calculate the new charge distribution, \(\sigma^\prime\)

Now, we will find the new charge distribution using the same formula, but with the new capacitance value, \(C^\prime\): \(\sigma^\prime = \frac{Q}{A}\) Use the new capacitance \(C^\prime\) value to calculate the new charge distribution, \(\sigma^\prime\).

b) Determine the potential difference \(V^\prime\)

When the battery is disconnected, the charge on the plates remains constant. We have to find the potential difference \(V^\prime\) between the plates when the distance is changed to \(0.250 \mathrm{~cm}\). Use the formula: \(V^\prime = \frac{Q}{C^\prime}\) Plug in the values of Q and \(C^\prime\) to calculate the potential difference, \(V^\prime\).

Key concepts

Capacitance Calculation

Understanding capacitance calculation is fundamental when dealing with electric circuits that include capacitors. Capacitance, denoted as 'C', measures a capacitor's ability to store charge per unit of electric potential difference. It’s expressed in farads (F), a large unit, so capacitance is often referenced in microfarads (\textmu F), nanofarads (nF), or picofarads (pF).

To calculate the capacitance using the relationship \(C = \frac{Q}{V}\), with 'C' being the capacitance, 'Q' the charge in coulombs, and 'V' the potential in volts, we must know two of these quantities. The example of a capacitor's capacitance changing when the distance between its plates is altered illustrates this concept well.

As distance between plates decreases, capacitance increases—this relationship is due to the fact that bringing charges closer together makes it easier to store more charge for a given voltage. In the exercise, we see the capacitance doubling when the distance is halved from 0.500 cm to 0.250 cm. This simple inverse relationship between distance and capacitance is crucial when designing circuits that need precise timing or filtering applications.

Charge Distribution on Plates

Charge distribution, represented by \(\sigma\), explains how charge is spread across the surface of the plates in a capacitor. A uniform \(\sigma\) is usually assumed for parallel-plate capacitors, meaning that the charge per unit area is the same across the entire plate surface.

In practice, we calculate \(\sigma\) using \(\sigma = \frac{Q}{A}\), where 'A' is the area of the plates and 'Q' is the total charge stored in the capacitor. The case where plate distance is modified without changing the total charge dramatically demonstrates that while the total charge 'Q' remains the same, the capacitance and thus the charge distribution will differ.

When the separation is reduced and the plates are closer, the same charge is spread over the same area but leads to a higher electric field between the plates. Consequently, understanding charge distribution is vital in designing capacitors, particularly in ensuring the electric field does not exceed the breakdown voltage of the material separating the plates.

Electric Potential Difference

The electric potential difference, commonly known as voltage and denoted by 'V', is the driving force that moves charge around in a circuit. It's the energy per unit charge created by separating positive and negative charges. A higher potential difference between two points means a greater ability to perform work, like moving charges through a conductor or storing energy in a capacitor.

In capacitors, the potential difference is directly proportional to the charge stored and inversely proportional to the capacitance (\(V = \frac{Q}{C}\)). This equation illustrates how potential difference changes with varying capacitance without altering the charge in the disconnection scenario of the exercise.

By decreasing the distance between capacitor plates from 0.500 cm to 0.250 cm after disconnecting the battery, the capacitance increased but the charge remained the same, hence the potential difference across the plates goes down. This concept has practical applications, such as adjusting the voltage stored in a camera flash capacitor.