Question

A parallel plate capacitor has plate of area \(\mathrm{A}\) and separation d. It is charged to a potential difference \(\mathrm{V}_{0}\). The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is (A) \(\left[\left(\mathrm{A} \in_{0} \mathrm{~V}_{0}^{2}\right) / \mathrm{d}\right]\) (B) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}^{2}\right) /(2 \mathrm{~d})\right]\) (C) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(3 \mathrm{~d})\right]\) (D) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]\)

Short answer

The work required to separate the plates is \(\left[\left(A \varepsilon_{0} V_{0}^{2}\right) /(2 d)\right]\).

Step-by-step solution

Capacitance Formula

Recall the capacitance formula of a parallel plate capacitor, which is given by \(C=\varepsilon_{0}\dfrac{A}{d}\), where \(\varepsilon_{0}\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the separation between the plates.

Potential Energy Stored in the Capacitor

The potential energy stored in the capacitor when it is charged to potential difference \(V_{0}\) is given by the formula \(U = \dfrac{1}{2} CV_{0}^{2}\). Using the capacitance formula, the potential energy can be expressed as \(U = \dfrac{1}{2}(\varepsilon_{0} \dfrac{A}{d})V_{0}^{2}\).

Increasing the Separation of the Plates

The plates are pulled apart to three times the initial separation. Let the new separation be \(d'\), then \(d' = 3d\). The capacitance after the separation is increased can be given by \(C' = \varepsilon_{0} \dfrac{A}{d'} = \varepsilon_{0} \dfrac{A}{3d}\).

Potential Energy at New Separation

Since the battery is disconnected from the capacitor, the charge on the plates is conserved and remains constant. Using the formula \(Q = CV\), and with constant \(Q\), the potential at new separation \(V'\) becomes \(V' = 3V_{0}\). Now, to find the potential energy stored in the capacitor at the new separation, we use the formula \(U' = \dfrac{1}{2} C'(V')^{2}\). Substituting the values of \(C'\) and \(V'\), we get \(U' = \dfrac{1}{2}(\varepsilon_{0} \dfrac{A}{3d}) (3V_{0})^{2}\).

Work Required to Separate the Plates

The work required to separate the plates is the difference between the potential energies at the new and initial separations, i.e., \(W = U' - U\). Plugging in the expressions for \(U'\) and \(U\), we get: \(W = \dfrac{1}{2}(\varepsilon_{0} \dfrac{A}{3d}) (3V_{0})^{2} - \dfrac{1}{2}(\varepsilon_{0} \dfrac{A}{d})V_{0}^{2}\) After simplifying, we get: \(W = \dfrac{1}{2} \varepsilon_{0} A(V_{0})^{2}\left(\dfrac{1}{d}-\dfrac{1}{3d}\right)\) Which simplifies further to: \(W = \dfrac{1}{2} \varepsilon_{0} A(V_{0})^{2}\left(\dfrac{2}{3d}\right)\) Comparing the expression of work with the given options, we find that the correct option is (B). Answer: The work required to separate the plates is \(\left[\left(A \varepsilon_{0} V_{0}^{2}\right) /(2 d)\right]\).

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor is a simple yet fundamental component in electronics and physics. It consists of two large, flat plates placed parallel to each other, separated by a small distance. These plates are typically made of metal, as metals are excellent conductors of electricity. The area of each plate is denoted by \( A \), and the distance between them is \( d \).

The key function of a capacitor is to store electrical energy in the form of an electrostatic field. When a voltage \( V \) is applied across the plates, electrons accumulate on one plate, creating a negative charge, while the other plate loses electrons and becomes positively charged. This setup allows the capacitor to hold energy even if the power source is removed. Moreover, the ability of a capacitor to store charge is determined by its capacitance, which is directly proportional to the plate area \( A \) and inversely proportional to the distance \( d \) between the plates.

Capacitance

Capacitance \( C \) is a measure of a capacitor's ability to store charge per unit voltage. It is expressed mathematically for a parallel plate capacitor as \( C = \varepsilon_0 \frac{A}{d} \). Here, \( \varepsilon_0 \) is the permittivity of free space, a constant that characterizes the ability of the vacuum to allow electric field lines to pass through it.

The larger the area \( A \), the more charge the capacitor can store, enhancing its capacitance. Conversely, increasing the distance \( d \) between the plates reduces the capacitance, as the electric field strength decreases with greater separation. Capacitors are essential in circuits for smoothing out voltage fluctuations, filtering signals, and storing small amounts of energy for quick release.

• Increasing plate area \( \rightarrow \) Increases capacitance
• Decreasing plate separation \( \rightarrow \) Increases capacitance
• Capacitance depends on the material between the plates (if any)

Electrostatic Potential Energy

Electrostatic potential energy is the energy stored within a capacitor due to the position of charges. For a capacitor charged to a potential difference \( V_0 \), the energy \( U \) is given by the expression \( U = \frac{1}{2} CV_0^2 \). This energy arises from the work done in moving charges against the electric field to establish this potential difference.

When the capacitor plates are pulled apart, as in the problem, the system's potential energy changes. Initially, the energy is higher because the electric field is stronger when the plates are closer. As the plates are separated without changing the charge, the potential energy at the new distance is different, reflecting the work required to separate the plates.

• Stored energy is due to charge separation
• Higher separation \( \rightarrow \) less energy stored per unit charge
• Energy can be converted to do work when released

Electrostatic Force

Electrostatic force is the force between charged objects due to their electric charges. In a parallel plate capacitor, this force is of attractive nature between the oppositely charged plates, pulling them together. This force is constant across the plates and depends on the amount of charge and the distance between them.

Bringing or separating charged capacitor plates involves work against this electrostatic force, which leads to a change in the stored energy, as discussed earlier. Calculating the work done, as in the given problem, involves accounting for the change in potential energy caused by this repositioning. The concept of electrostatic force is crucial in understanding how energy and forces interact in electric fields.

• Force varies with the square of charge
• Inverse relation with the square of distance
• Responsible for holding charges onto the plates