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(II) \((a)\) Show that each plate of a parallel-plate capacitor exerts a force $$F=\frac{1}{2} \frac{Q^{2}}{\epsilon_{0} A}$$ on the other, by calculating dW/dx where \(d W\) is the work needed to increase the separation by \(d x\) . (b) Why does using \(F=Q E,\) with \(E\) being the electric field between the plates, give the wrong answer?

Short Answer

Expert verified
(a) The force is \(F = \frac{1}{2} \frac{Q^2}{\epsilon_0 A}\). (b) \(F=QE\) overlooks work done during separation change.

Step by step solution

01

Understand the Problem

We need to find the force between plates of a parallel-plate capacitor using the concept of work done in increasing the separation between the plates.
02

Use Capacitor Energy Formula

The energy stored in a capacitor is given by \( U = \frac{1}{2} \frac{Q^2}{C} \), where \( C = \frac{\epsilon_0 A}{d} \). As the separation \( d \) changes by \( dx \), the new capacitance becomes crucial for further calculation.
03

Calculate dW, the Change in Energy

The change in energy \( dW \) when the capacitor's separation increases by \( dx \) is \( dU = \frac{d}{dx}\left(\frac{1}{2} \frac{Q^2}{C}\right)\cdot dx \). This will involve differentiating the energy with respect to distance.
04

Differentiate the Energy Expression

Differentiate \( U = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} \frac{Q^2 d}{\epsilon_0 A} \) with respect to \( x \). This gives \( dU = \frac{1}{2} \frac{Q^2}{\epsilon_0 A} \cdot dx \).
05

State dW/dx as Force

The work done \( dW \) to increase the plate separation by \( dx \) is equal to the force \( F \) times the distance, so \( F = \frac{dW}{dx} = \frac{1}{2} \frac{Q^2}{\epsilon_0 A} \). This shows the force exerted by each plate.
06

Explain Why F = QE is Incorrect

Using \( F = Q E \) considers electric field \( E = \frac{Q}{\epsilon_0 A} \) for an isolated charge. However, for capacitors, the energy consideration due to electric field distribution between plates gives correct force.

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Key Concepts

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

Capacitor Energy Formula
A capacitor is a device that stores electrical energy through the separation of charges. The energy stored in a capacitor, given a charge \( Q \) on it, can be calculated using the formula: \[ U = \frac{1}{2} \frac{Q^2}{C} \] where \( U \) is the energy, \( Q \) is the charge, and \( C \) is the capacitance of the capacitor.
The capacitance \( C \) for a parallel-plate capacitor, which consists of two plates separated by a distance \( d \) and having an overlapping area \( A \), is \[ C = \frac{\epsilon_0 A}{d} \] where \( \epsilon_0 \) is the permittivity of free space. As the distance \( d \) changes, the capacitance \( C \) also changes, affecting the stored energy. Understanding how energy in a capacitor changes with distance is crucial when calculating forces in such a system.
Work and Energy in Physics
In the context of a parallel-plate capacitor, work and energy concepts are key to understanding forces between the plates. When the separation \( dx \) between the plates changes, the work done is related to the change in stored energy in the capacitor.
The work done \( dW \) to change the plate separation by \( dx \) can be expressed as: \[ dW = F \cdot dx \] where \( F \) is the force exerted by one plate on the other. In this scenario, the change in energy \( dU \) is crucial for determining \( F \). By differentiating the energy formula in terms of the separation, we get: \[ dU = \frac{d}{dx}\left(\frac{1}{2} \frac{Q^2}{C}\right)\cdot dx \] This leads to the expression for force: \[ F = \frac{1}{2} \frac{Q^2}{\epsilon_0 A} \] reflecting how energy and force are intertwined in a nuanced way in capacitors.
Electric Field in Capacitors
The electric field in a parallel-plate capacitor is essential for understanding forces. It is typically uniform between the plates and calculated as: \[ E = \frac{V}{d} \] where \( V \) is the voltage. For a charge \( Q \) and capacitance \( C \), it can also be written as: \[ E = \frac{Q}{\epsilon_0 A} \] However, using \( F = Q E \) to find the force between capacitor plates is incorrect in this context. This common mistake overlooks the energy distribution's role.
  • The electric field \( E \) above primarily considers isolated charges, not the mutual influence in capacitors.
  • Capacitor plates both store and affect energy, meaning accurate force calculations must consider changing energy levels.
The formula \( F = \frac{1}{2} \frac{Q^2}{\epsilon_0 A} \) derived from energy changes accurately reflects these intricacies, providing the correct assessment of the forces involved.

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Most popular questions from this chapter

Let us try to estimate the maximum "static electricity" charge that might result during each walking step across an insulating floor. Assume the sole of a person's shoe has area \(A \approx 150 \mathrm{~cm}^{2},\) and when the foot is lifted from the ground during each step, the sole acquires an excess charge \(Q\) from rubbing contact with the floor. (a) Model the sole as a plane conducting surface with \(Q\) uniformly distributed across it as the foot is lifted from the ground. If the dielectric strength of the air between the sole and floor as the foot is lifted is \(E_{\mathrm{S}}=3 \times 10^{6} \mathrm{~N} / \mathrm{C},\) determine \(Q_{\max },\) the maximum possible excess charge that can be transferred to the sole during each step. (b) Modeling a person as an isolated conducting sphere of radius \(r \approx 1 \mathrm{~m},\) estimate a person's capacitance. \((c)\) After lifting the foot from the floor, assume the excess charge \(Q\) quickly redistributes itself over the entire surface area of the person. Estimate the maximum potential difference that the person can develop with respect to the floor.

A parallel-plate capacitor is isolated with a charge \(\pm Q\) on each plate. If the separation of the plates is halved and a dielectric (constant \(K\) ) is inserted in place of air, by what factor does the energy storage change? To what do you attribute the change in stored potential energy? How does the new value of the electric field between the plates compare with the original value?

(1) 2200 \(\mathrm{V}\) is applied to a \(2800-\mathrm{pF}\) capacitor. How much electric energy is stored?

(II) A \(7.7-\mu \mathrm{F}\) capacitor is charged by a \(125-\mathrm{V}\) battery (Fig. 20 \(\mathrm{a} )\) and then is disconnected from the battery. When this capacitor \(\left(C_{1}\right)\) is then connected (Fig. 20 \(\mathrm{b} )\) to a second (initially uncharged) capacitor, \(C_{2},\) the final voltage on each capacitor is 15 \(\mathrm{V}\) . What is the value of \(C_{2} ?[\)Hint. . Charge is conserved.]

In an electrostatic air cleaner ( "precipitator"), the strong nonuniform electric field in the central region of a cylindrical capacitor (with outer and inner cylindrical radii \(R_{\mathrm{a}}\) and \(R_{\mathrm{b}}\) ) is used to create ionized air molecules for use in charging dust and soot particles (Fig. \(24-21\) ). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength \(E_{\mathrm{S}}=2.7 \times 10^{6} \mathrm{~N} / \mathrm{C},\) air molecules will dissociate into positively charged ions and free electrons. In a precipitator, the region within which air is ionized (the corona discharge region) occupies a cylindrical volume of radius \(R\) that is typically five times that of the inner cylinder. Assume a particular precipitator is constructed with \(R_{\mathrm{b}}=0.10 \mathrm{~mm}\) and \(R_{\mathrm{a}}=10.0 \mathrm{~cm} .\) In order to create a corona discharge region with radius \(R=5.0 \mathrm{R}_{\mathrm{b}},\) what potential difference \(V\) should be applied between the precipitator's inner and outer conducting cylinders? [Besides dissociating air, the charged inner cylinder repels the resulting positive ions from the corona discharge region, where they are put to use in charging dust particles, which are then "collected" on the negatively charged outer cylinder.]

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