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Two large metal parallel plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{~mm},\) and the potential difference between them is \(360 \mathrm{~V}\). (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+2.40 \mathrm{nC} ?\)

Short Answer

Expert verified
(a) Electric field: 8000 V/m. (b) Force: 1.92 x 10^-5 N.

Step by step solution

01

Understand the Relationship Between Potential Difference and Electric Field

The electric field \( E \) between two parallel plates can be calculated using the formula \( E = \frac{V}{d} \), where \( V \) is the potential difference, and \( d \) is the separation between the plates.
02

Convert Units as Necessary

The separation between the plates is given as \( 45.0 \, \text{mm} \). Convert this into meters to match the potential difference units, \( d = 45.0 \, \text{mm} = 0.045 \, \text{m} \).
03

Calculate the Electric Field

Substitute the given values into the formula: \( E = \frac{360 \, \text{V}}{0.045 \, \text{m}} \). Calculate this to find the magnitude of the electric field.
04

Perform the Calculation for the Electric Field

Calculate \( E = \frac{360}{0.045} = 8000 \, \text{V/m} \). So, the electric field magnitude between the plates is \( 8000 \, \text{V/m} \).
05

Understand the Force on a Charged Particle in an Electric Field

The force \( F \) on a particle in an electric field is calculated using the formula \( F = qE \), where \( q \) is the charge of the particle and \( E \) is the electric field.
06

Convert the Particle's Charge to Coulombs

The charge given is \( +2.40 \, \text{nC} \). Convert this into Coulombs: \( q = 2.40 \, \text{nC} = 2.40 \times 10^{-9} \, \text{C} \).
07

Calculate the Force on the Particle

Substitute the values into the formula: \( F = (2.40 \times 10^{-9} \, \text{C})(8000 \, \text{V/m}) \). Calculate this to find the magnitude of the force.
08

Perform the Calculation for the Force

Calculate \( F = 2.40 \times 10^{-9} \times 8000 = 1.92 \times 10^{-5} \, \text{N} \). So, the force exerted on the particle is \( 1.92 \times 10^{-5} \, \text{N} \).

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Key Concepts

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

Potential Difference
Potential difference, often referred to as voltage, is a measure of the work needed to move a charge between two points in an electric field. It is crucial in understanding how electric fields behave between
  • charged objects,
  • power sources, or
  • within circuits.
The potential difference is what causes charged particles to move and is measured in volts (V). In the context of parallel plates, it represents the difference in electric potential energy per unit charge between the plates. For example, if the potential difference between two large metal plates is 360 V, it signifies that 360 joules of work is needed to move one coulomb of charge from one plate to the other. This potential difference directly impacts the electric field generated between the plates.
Parallel Plates
Parallel plates are used to create uniform electric fields between them. They are two conducting surfaces
  • placed at a distance from each other
  • and connected to a voltage source.
The plates carry equal and opposite charges, resulting in an electric field that is constant in magnitude.
The electric field (E) between two parallel plates is directly related to the potential difference (V) and the distance (d) between the plates, described by the equation:\[E = \frac{V}{d}\]If the separation distance is 0.045 m and the potential difference is 360 V, this equation enables us to calculate the electric field, which in this case is 8000 V/m. This setup is widely used in experiments and technology to achieve controllable electric fields.
Charged Particle
A charged particle in an electric field experiences forces that depend on its charge magnitude and the field strength. The charge of a particle is usually given in terms of coulombs (C), but it might also be in nanocoulombs (nC), where 1 nC = \( 10^{-9} \) C. In the given exercise, the charge is 2.40 nC. The interaction of the charged particle with an electric field can be easily understood using the equation \( F = qE \), where:
  • \( F \) is the force exerted on the charge,
  • \( q \) is the charge of the particle,
  • \( E \) is the electric field strength.
This force determines how the particle moves within the field and is fundamental in analyzing particle behavior in experimental and real-world applications.
Force Calculation
Calculating the force on a charged particle within an electric field is essential to understanding its motion and interactions. Using the force equation,\[F = qE\],you can determine the force acting on a particle as long as you know its charge and the electric field.

For example, with a charge of 2.40 nC (or 2.40 × \( 10^{-9} \) C) and an electric field of 8000 V/m, the force calculation becomes straightforward:\[F = (2.40 \times 10^{-9} \, \text{C}) \times 8000 \, \text{V/m} = 1.92 \times 10^{-5} \, \text{N}\]This process allows you to predict how charged particles will react to electric fields, which is crucial in fields such as electronics, physics, and engineering.

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Most popular questions from this chapter

\(\mathrm{A} 12.5 \mu \mathrm{F}\) capacitor is connected to a power supply that keeps a constant potential difference of \(24.0 \mathrm{~V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

An alpha particle with a kinetic energy of \(10.0 \mathrm{MeV}\) makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is \(79,\) and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)

A helium ion \(\left(\mathrm{He}^{++}\right)\) that comes within about \(10 \mathrm{fm}\) of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of \(3.0 \mathrm{MeV}\) heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than \(10 \mathrm{fm}\) from the center of the atomic nucleus? (1 \(\mathrm{fm}=1 \times 10^{-15} \mathrm{~m},\) and \(e\) is the magnitude of the charge of an electron or a proton. A. \(2 e\) B. \(11 e\) C. \(20 e\) D. \(22 e\)

A parallel-plate capacitor having plates \(6.0 \mathrm{~cm}\) apart is connected across the terminals of a \(12 \mathrm{~V}\) battery. (a) Being as quantitative as you can, describe the location and shape of the equipotential surface that is at a potential of \(+6.0 \mathrm{~V}\) relative to the potential of the negative plate. Avoid the edges of the plates. (b) Do the same for the equipotential surface that is at \(+2.0 \mathrm{~V}\) relative to the negative plate. (c) What is the potential gradient between the plates?

A small particle has charge \(-5.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-4} \mathrm{~kg}\). It moves from point \(A\), where the electric potential is \(V_{A}=+200 \mathrm{~V}\), to point \(B\), where the electric potential is \(V_{B}=+800 \mathrm{~V}\). The electric force is the only force acting on the particle. The particle has speed \(5.00 \mathrm{~m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A\) ? Explain.

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