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Two large, parallel, conducting plates are \(12 \mathrm{~cm}\) apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electric force of \(3.9 \times 10^{-15} \mathrm{~N}\) acts on an electron placed anywhere between the two plates. (Neglect fringing.) (a) Find the electric field at the position of the electron. (b) What is the potential difference between the plates?

Short Answer

Expert verified
Electric field: 24,400 N/C; Potential difference: 2,930 V.

Step by step solution

01

Understand the Problem

We have two parallel conducting plates with an electron placed between them. The problem provides the force acting on the electron and requires the calculation of the electric field at the electron’s location and the potential difference between the plates.
02

Calculate the Electric Field

The electric force acting on a charge in an electric field is given by the equation: \( F = qE \), where \( F \) is force, \( q \) is the charge, and \( E \) is the electric field. For an electron, the charge \( q = -1.6 \times 10^{-19} \mathrm{~C} \). Given \( F = 3.9 \times 10^{-15} \mathrm{~N} \), we can solve for \( E \): \[ E = \frac{F}{q} = \frac{3.9 \times 10^{-15} \mathrm{~N}}{-1.6 \times 10^{-19} \mathrm{~C}}. \] After calculating, we find the magnitude of the electric field: \( E = 2.44 \times 10^{4} \mathrm{~N/C} \).
03

Calculate the Potential Difference

The potential difference \( V \) between two plates is related to the electric field \( E \) by the equation \( V = Ed \), where \( d \) is the distance between the plates. Given \( d = 12 \mathrm{~cm} = 0.12 \mathrm{~m} \) and \( E = 2.44 \times 10^{4} \mathrm{~N/C} \), we calculate: \[ V = (2.44 \times 10^{4} \mathrm{~N/C})(0.12 \mathrm{~m}) = 2.928 \times 10^{3} \mathrm{~V}. \] Thus, the potential difference is \( 2.93 \mathrm{~kV} \).

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Key Concepts

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

Electric Force
Electric force is one of the fundamental forces of nature. It is the force experienced by a charged particle in an electric field.
To calculate the electric force (\( F \)) acting on a charge (\( q \)), we use:
  • \( F = qE \).
The electric force is crucial because it defines how charged particles interact with each other and their surroundings.
In the context of the problem, an electric force of \( 3.9 \times 10^{-15} \, \mathrm{N} \) acts on an electron.
Given the charge of an electron is negative, \( q = -1.6 \times 10^{-19} \, \mathrm{C} \), we can determine the electric field (\( E \)) by rearranging the formula:
  • \( E = \frac{F}{q} \).
Electric forces not only cause charged particles to move but also determine how they balance within electric fields, like between two conducting plates.
This understanding allows us to determine both the magnitude and direction of such interactions.
Potential Difference
Potential difference, often referred to as voltage, is the change in electric potential energy per unit charge between two points in an electric field.
In simpler terms, it tells us how much "push" an electric field has between two points.
The relationship between potential difference (\( V \)), electric field (\( E \)), and distance (\( d \)) is captured by:
  • \( V = Ed \).
This formula is derived from the definition of potential difference and the work done in moving a charge against an electric field.
For parallel plates, the potential difference is consistent across the field created between the plates.
In this scenario, with the plates being \( 0.12 \mathrm{~m} \) apart, and knowing the electric field is \( 2.44 \times 10^{4} \, \mathrm{N/C} \), the potential difference calculates as:
  • \( V = 2.928 \times 10^{3} \, \mathrm{V} \),
which is equivalent to \( 2.93 \, \mathrm{kV} \).
This depicts the energy needed to move a charge from one plate to the other.
Conducting Plates
Conducting plates are materials that allow electric charges to move across them easily.
In physics, particularly electrostatics, the concept of parallel conducting plates is crucial.
When two large, conducting plates are placed parallel to each other and charged, they create a uniform electric field in the space between them.
This occurs because charges redistribute on the plates’ surfaces to ensure the electric field within the conductor is zero.
The configuration:
  • ensures a consistent electric field between the plates,
  • prevents external electric fields from influencing the charges within,
  • creates an environment where calculations like potential difference and electric field strength are straightforward.
The uniformly distributed electric field means any charge placed between the plates experiences a force of constant magnitude and direction.
This characteristic is utilized in technological applications like capacitors, where such plate configurations can store and release energy efficiently.

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Most popular questions from this chapter

The magnitude \(E\) of an electric field depends on the radial distance \(r\) according to \(E=A / r^{4}\), where \(A\) is a constant with the unit volt-cubic meter. As a multiple of \(\bar{A}\), what is the magnitude of the electric potential difference between \(r=2.00 \mathrm{~m}\) and \(r=3.00 \mathrm{~m} ?\)

Two isolated, concentric, conducting spherical shells have radii \(R_{1}=0.500 \mathrm{~m}\) and \(R_{2}=1.00 \mathrm{~m}\), uniform charges \(q_{1}=+2.00 \mu \mathrm{C}\) and \(q_{2}=+1.00 \mu \mathrm{C}\), and negligible thicknesses. What is the magnitude of the electric field \(E\) at radial distance (a) \(r=4.00 \mathrm{~m},(\mathrm{~b}) r=\) \(0.700 \mathrm{~m}\), and (c) \(r=0.200 \mathrm{~m}\) ? With \(V=0\) at infinity, what is \(V\) at (d) \(r=4.00 \mathrm{~m}\), (e) \(r=1.00 \mathrm{~m}\), (f) \(r=0.700 \mathrm{~m}\), (g) \(r=0.500 \mathrm{~m}\), (h) \(r=0.200 \mathrm{~m}\), and (i) \(r=0 ?\) (j) \(\operatorname{Sketch} E(r)\) and \(V(r)\).

A particle of positive charge \(Q\) is fixed at point \(P .\) A second particle of mass \(m\) and negative charge \(-q\) moves at constant speed in a circle of radius \(r_{1}\), centered at \(P .\) Derive an expression for the work \(W\) that must be done by an external agent on the second particle to increase the radius of the circle of motion to \(r_{2}\).

A particle of charge \(+7.5 \mu \mathrm{C}\) is released from rest at the point \(x=60 \mathrm{~cm}\) on an \(x\) axis. The particle begins to move due to the presence of a charge \(Q\) that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved \(40 \mathrm{~cm}\) if (a) \(Q=+20 \mu \mathrm{C}\) and (b) \(Q=-20 \mu \mathrm{C}\) ?

A spherical drop of water carrying a charge of 30 \(\mathrm{pC}\) has a potential of \(500 \mathrm{~V}\) at its surface (with \(V=0\) at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

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