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Chapter 1: Problem 29

Pulling two sheets apart ** Two parallel sheets each have large area \(A\) and are separated by a small distance \(\ell\). The surface charge densities are \(\sigma\) and \(-\sigma\). You wish to pull one of the sheets away from the other, by a small distance \(x\). How much work does this require? Calculate this by: (a) using the relation \(W=\) (force) \(\times\) (distance); (b) calculating the increase in energy stored in the electric field. Show that these two methods give the same result.

Short Answer

Expert verified
The work required to pull the sheets apart by a small distance \(x\) is \( (\sigma^2/蔚_0) A x \). Both the methods of (a) force \( \times \) distance and (b) change in stored electric field energy give the same result.

Step by step solution

01

Understanding electrostatic forces

An attractive force exists between two surfaces having opposite charges. This force depends on the electric field \(E\) between the sheets. The electric field between oppositely charged parallel sheets is \(E = \sigma/蔚_0\), where \(蔚_0\) is the permittivity of free space. Therefore, the force per unit area is \(F/A = E \times \sigma = \sigma^2/蔚_0\).
02

Calculate work done via force 脳 distance

Work \(W\) is defined as force \(F\) times distance \(d\). In this case, the work \(W\) to pull one sheet away by a small distance \(x\) is \(W = F \times x = (F/A) \times A \times x = (\sigma^2/蔚_0) \times A \times x\).
03

Calculate increase in stored energy

The energy stored in an electric field is given by the formula \(U = 1/2 蔚_0 E^2\). Hence the energy per unit volume in this case is \(1/2 蔚_0 E^2 = 1/2 蔚_0 (\sigma/蔚_0)^2 = 1/2 \sigma^2/蔚_0\). As \(U\) increases with increasing separation, we calculate increase in energy by integrating this energy density from \(\ell\) to \( \ell + x\), \(螖 U = 鈭玙 \ell ^{\ell+x} (1/2) (\sigma^2/蔚_0) dy = 1/2 (\sigma^2/蔚_0) x\). The increase in energy for area \(A\) is \(螖 W = A (1/2) (\sigma^2/蔚_0) x\).
04

Compare results from both methods

Comparing the results from step 2 (\(W = (\sigma^2/蔚_0) A x\)) and step 3 (\(螖 W = A (1/2) (\sigma^2/蔚_0) x\)), we can see that they are same, thereby proving that both methods give the same result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrostatic Forces
Electrostatic forces are fundamental interactions between charged particles. These forces are described by Coulomb's law, which states that two point charges exert a force on each other that is proportional to the product of their charges and inversely proportional to the square of the distance between them. In our example with two parallel sheets of opposite charge, we simplify this concept as we deal with uniform surface charge densities over large areas.

The expression for the electric field that arises between two oppositely charged plates, given by the equation
\( E = \frac{\sigma}{\varepsilon_0} \),
allows us to understand the uniform force acting over the area of the plates. The attractive electrostatic force per unit area between the sheets can then be calculated as
\( F/A = E \times \sigma = \frac{\sigma^2}{\varepsilon_0} \).
This formula is crucial for understanding how work is done when the plates are pulled apart, translating the electrostatic interaction into a mechanical process of doing work.
Work-Energy Principle in Electrostatics
The work-energy principle is a foundational concept in physics that links the work done on an object to the change in its energy. In electrostatics, this principle helps us understand how the mechanical work required to move charged objects translates into changes in the electrostatic energy of a system. In our exercise involving pulling apart two charge sheets, we calculate the work done using the equation
\( W = F \times x \),
where
\( F \)
represents the total electrostatic force exerted on one sheet due to the presence of the other, and
\( x \)
is the distance over which the sheet is moved.

Calculating the work done in moving the sheet increases the separation between the sheets and therefore the potential energy stored in the electric field. This increase corresponds to an increase in the electrostatic energy of the system. The work-energy principle ensures that the work done on the system is equal to the change in its energy, which is verified by the exercise's result showing the consistency between the force-distance method and the energy storage method.
Permittivity of Free Space
The permittivity of free space, denoted by
\( \varepsilon_0 \),
is a fundamental physical constant that describes how electric fields interact with the vacuum of space. It plays a pivotal role in the equations governing electrostatic phenomena. For example, the permittivity of free space appears in Coulomb's law as a scaling factor that affects the strength of the electrostatic force between charges in vacuum.

In our scenario where we calculate the force between parallel charged plates,
\( \varepsilon_0 \)
is essential to determine the electric field generated by a given surface charge density using
\( E = \frac{\sigma}{\varepsilon_0} \).
It also plays a key role in determining the stored electrostatic energy in the electric field between the sheets. A higher permittivity allows the field to store more energy for a given electric field strength, which in the context of our exercise would mean more work is required to separate the charged sheets. This underscores the contribution of
\( \varepsilon_0 \)
to the energy relationships in electrostatic systems, such as capacitors and other electric field configurations.

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Most popular questions from this chapter

Maximum field from a ring ** A charge \(Q\) is distributed uniformly around a thin ring of radius \(b\) that lies in the \(x y\) plane with its center at the origin. Locate the point on the positive \(z\) axis where the electric field is strongest.

(a) Two rings with radius \(r\) have charge \(Q\) and \(-Q\) uniformly distributed around them. The rings are parallel and located a distance \(h\) apart, as shown in Fig. \(1.35\). Let \(z\) be the vertical coordinate, with \(z=0\) taken to be at the center of the lower ring. As a function of \(z\), what is the electric field at points on the axis of the rings? (b) You should find that the electric field is an even function with respect to the \(z=h / 2\) point midway between the rings. This implies that, at this point, the field has a local extremum as a function of \(z\). The field is therefore fairly uniform there; there are no variations to first order in the distance along the axis from the midpoint. What should \(r\) be in terms of \(h\) so that the field is very uniform? By "very" uniform we mean that additionally there aren't any variations to second order in \(z\). That is, the second derivative vanishes. This then implies that the leading-order change is fourth order in \(z\) (because there are no variations at any odd order, since the field is an even function around the midpoint). Feel free to calculate the derivatives with a computer.

\(N\) charges on a circle \(N\) point charges, each with charge \(Q / N\), are evenly distributed around a circle of radius \(R\). What is the electric field at the location of one of the charges, due to all the others? (You can leave, your answer in the form of a sum.) In the \(N \rightarrow \infty\) limit, is the field infinite or finite? In the \(N \rightarrow \infty\) limit, is the force on one of the charges infinite or finite?

\mathrm{~ F o r c e s ~ o n ~ t h r e e ~ s h e e t s ~ * ?}\( Consider three charged sheets, \)A, B\(, and \)C .\( The sheets are parallel with \)A\( above \)B\( above \)C\(. On each sheet there is surface charge of uniform density: \)-4 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\( on \)A, 7 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\( on \)B\(, and \)-3 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\( on \)C$. (The density given includes charge on both sides of the sheet.) What is the magnitude of the electrical force per unit area on each sheet? Check to see that the total force per unit area on the three sheets is zero.

Force on a patch Consider a small patch of charge that is part of a larger surface. The surface charge density is \(\sigma .\) If \(\mathbf{E}_{1}\) and \(\mathbf{E}_{2}\) are the electric fields on either side of the patch, show that the force per unit area on the patch equals \(\sigma\left(\mathbf{E}_{1}+\mathbf{E}_{2}\right) / 2\). This is the result we derived in Section \(1.14\), for the case where the field is perpendicular to the surface. Derive it here by using the fact that the force on the patch is due to the field \(\mathbf{E}^{\text {other }}\) from all the other charges in the system (excluding the patch), and then finding an expression for \(\mathbf{E}^{\text {other }}\) in terms of \(\mathbf{E}_{1}\) and \(\mathbf{E}_{2}\)
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