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A thin conducing plate \(2.0 \mathrm{m}\) on a side is given a total charge of \(-10.0 \mu \mathrm{C}\). (a) What is the electric field \(1.0 \mathrm{cm}\) above the plate? (b) What is the force on an electron at this point? (c) Repeat these calculations for a point \(2.0 \mathrm{cm}\) above the plate. (d) When the electron moves from 1.0 to \(2,0 \mathrm{cm}\) above the plate, how much work is done on it by the electric field?

Short Answer

Expert verified
The electric field at both 1.0 cm and 2.0 cm above the plate is -70.6 N/C. The force on an electron at these points is 1.13 x 10^{-17} N, and the work done by the electric field when the electron moves from 1.0 cm to 2.0 cm above the plate is 1.13 x 10^{-18} J.

Step by step solution

01

Calculate the charge density on the plate

To find the electric field due to the conducting plate, we first need to calculate the charge density on the plate. The charge density (σ) is the total charge (Q) divided by the area (A) of the plate. In this case, the total charge is -10.0 μC and the plate is a square with dimensions 2.0 m x 2.0 m. σ = \(\frac{Q}{A}\) σ = \(\frac{-10.0 \times 10^{-6} C}{(2.0m)(2.0m)}\) = \(-1.25 \times 10^{-6} C/m^2\)
02

Calculate the electric field above the plate

To calculate the electric field (E) above a charged plate at distance d from the plate, we can use the following formula: E = \(\frac{σ}{2ε_0}\) where ε_0 is the vacuum permittivity constant, approximately \(8.85 \times 10^{-12} C^2/Nm^2\). (a) For 1.0 cm above the plate: d = 1.0 cm = 0.01 m E1 = \(\frac{-1.25\times 10^{-6} C/m^2}{2(8.85\times 10^{-12} C^{2}/Nm^{2})}\) = -70.6 N/C (b) For 2.0 cm above the plate: d = 2.0 cm = 0.02 m E2 = \(\frac{-1.25\times 10^{-6} C/m^2}{2(8.85\times 10^{-12} C^{2}/Nm^{2})}\) = -70.6 N/C Note that the electric field is the same at both distances because it is a thin plate.
03

Calculate the force on an electron at each point

To calculate the force on an electron (F) due to the electric field, we can use the following formula: F = qE where q is the charge of an electron, approximately -1.60 x 10^{-19} C. (a) Force at 1.0 cm: F1 = (-1.60 x 10^{-19} C)(-70.6 N/C) = 1.13 x 10^{-17} N (b) Force at 2.0 cm: F2 = (-1.60 x 10^{-19} C)(-70.6 N/C) = 1.13 x 10^{-17} N Note that the force is the same at both distances as the electric field is the same.
04

Calculate the work done on the electron by the electric field

(d) When the electron moves from 1.0 cm to 2.0 cm above the plate, the electric field is constant, and the distance moved by the electron (Δd) is 0.01 m. To calculate the work done (W) on the electron by the electric field, we can use the following formula: W = FΔd W = (1.13 x 10^{-17} N)(0.01 m) = 1.13 x 10^{-18} J The work done on the electron by the electric field is 1.13 x 10^{-18} J.

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Key Concepts

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

Understanding Charge Density
When delving into the world of electromagnetism, charge density plays a pivotal role in understanding how electric fields behave. It's a measure of how much electric charge is accumulated in a particular area. Imagine spreading a layer of tiny charged particles over a surface; the density of that layer would indicate how 'crowded' the surface is with charges. To put it simply, charge density is the electric charge per unit area.

In practical scenarios, such as a conducting plate, you would determine charge density by dividing the total charge by the surface area of the plate. This concept isn't just a theoretical idea; it's essential for predicting how the electric field emanates from charged surfaces. A higher charge density means a stronger electric field close to the surface. It's also worth noting that the charge density can be positive or negative, depending on the type of charge that is present on the object.
Force on an Electron
When you place an electron—a tiny particle with a negative charge—near other charges, it will experience a force. This force is a direct consequence of the electric field generated by those charges. The force on an electron can be calculated using the formula F = qE, where F is the force, q is the charge of the electron, and E is the electric field strength.

An electron in the vicinity of a charged plate feels the force because of the electric field produced by the plate's charge. Even a small electron, with its minuscule charge, will experience this force. In our exercise, regardless of the distance from the plate (within a certain limit), the force remains constant because the electric field is uniform above a charged, flat surface. This uniformity is typical for an infinitely large plate, but can be applied to finite plates if the distance from the plate is relatively small, which is a useful approximation in many scenarios.
Work Done by Electric Field
Finally, let's talk about the work done by an electric field. Work can be understood as the energy transferred when a force moves an object over a distance. In the case of an electron moving in an electric field, the work done is the force multiplied by the distance moved.

Since the electron carries a charge, and it's moving through an area influenced by the electric field, this field will do work on the electron. Calculating this work can tell you how much energy was transferred to the electron as it moved. The formula used is W = FΔd. It's fascinating to note that if the electric field is constant, as it was when we looked at the plate above the electron at varying heights, the work done is simply the product of the force and the distance moved. The direction of the force relative to the movement of the electron is important too; if the electron moves against the direction of the force, the work is considered positive, indicating that energy is being consumed to move the electron.

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Most popular questions from this chapter

Suppose someone tells you that rubbing quartz with cotton cloth produces a third kind of charge on the quartz. Describe what you might do to test this claim.

Two thin parallel conducting plates are placed 2.0 \(\mathrm{cm}\) apart. Each plate is \(2.0 \mathrm{cm}\) on a side; one plate carries a net charge of \(8.0 \mu \mathrm{C}, \quad\) and the other plate carries a net charge of \(-8.0 \mu \mathrm{C} .\) What is the charge density on the inside surface of each plate? What is the electric field between the plates?

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