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A doubly charged ion is accelerated to an energy of \(32.0 \mathrm{keV}\) by the electric field between two parallel conducting plates separated by \(2.00 \mathrm{~cm}\). What is the electric field strength between the plates?

Short Answer

Expert verified
The electric field strength between the plates is 800 kV/m.

Step by step solution

01

Understanding the Concept

The potential energy (PE) gained by a charged particle in an electric field is given by PE = qV, where q is the charge of the particle and V the potential difference (voltage). When a charged ion is accelerated by an electric field, its kinetic energy (KE) must equal the potential energy it gains. Also, electric field strength E is the potential difference V divided by the distance d between the plates: E = V/d.
02

Calculating the Potential Difference

First, convert the kinetic energy from keV to joules (J) since 1 eV = 1.602 x 10^-19 J. Given that KE = 32.0 keV, then KE = 32.0 x 10^3 eV x 1.602 x 10^-19 J/eV = 5.1264 x 10^-15 J. The kinetic energy gained is equal to the change in potential energy, so V = KE/q, where q = 2e because the ion is doubly charged. Hence, V = (5.1264 x 10^-15 J) / (2 x 1.602 x 10^-19 C) = 16,000 volts.
03

Determining the Electric Field Strength

Finally, compute the electric field strength E using the equation E = V/d, where d = 2.00 cm = 0.02 m. So, E = (16,000 V) / (0.02 m) = 800,000 V/m or 800 kV/m.

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Key Concepts

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

Potential Energy
Potential energy (PE) is stored energy that is associated with the position of an object within a force field, such as the gravitational field or an electric field. In the context of charged particles, PE is the energy that a charge possesses due to its position in an electric field.

The formula PE = qV represents the potential energy of a charge in an electric field, where 'q' denotes the charge, and 'V' signifies the potential difference (voltage) across the field. Consider an example where an electron is lifted within an electric field; as it moves against the field's direction, it gains potential energy. This potential energy can be converted into kinetic energy if the electron is allowed to accelerate back through the field.
Kinetic Energy
Kinetic energy (KE) is the energy that an object possesses due to its motion. Unveiling the relation between kinetic energy and charges in electric fields, a charged particle accelerates when subjected to an electric field, thereby converting its potential energy into kinetic energy.

In our textbook exercise, the charged ion's kinetic energy is equal to the potential energy it acquired from the electric field, as indicated by the principle of energy conservation. If we know the kinetic energy a particle has acquired, we can deduce the potential energy it must have had before being accelerated.
Electric Field
An electric field is a vector field that surrounds electric charges, exerting a force on other charges within the field. The strength of this invisible field determines how strongly a charge would be acted upon. It is often visualized with field lines spreading out from a positive charge or converging onto a negative charge.

Electric Field Direction

The direction of the electric field is defined as the direction of the force on a positive test charge. Accordingly, the field lines are directed away from positive charges and toward negative charges.

Electric Field Intensity

The field's intensity is quantified as force per unit charge, with the SI unit being volts per meter (V/m). As the force acting on a charge increases, or as the charge's distance from the source charge decreases, the electric field strength similarly increases.
Charge
Charge is an intrinsic property of subatomic particles that causes them to experience a force when placed in an electric or magnetic field. Charges come in two varieties—positive and negative. Like charges repel each other, while unlike charges attract each other.

Charges are measured in coulombs (C), and the force between charges is described by Coulomb's law. In our problem, we encounter a doubly charged ion, which implies that the ion carries twice the elementary charge of an electron (approximately 1.602 x 10^-19 C). The interaction of this charge with the electric field between the plates is what propels the ion and alters its energy states.
Potential Difference
Potential difference, also known as voltage, represents the work done per unit charge to move a charge from one point to another within an electric field. It is the driving 'force' that facilitates the movement of charges in a circuit or between two points in space.

The potential difference is what we calculate when we want to determine the change in potential energy per unit charge for an electric field. In our scenario, this voltage propels the doubly charged ion from a state of lower potential energy to a state of higher kinetic energy as it accelerates between the parallel conducting plates. The relationship voltage (V) = energy (E) / charge (q) is pivotal for understanding how electric fields do work on charges.

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Most popular questions from this chapter

Suppose you want a capacitor bank with a total capacitance of \(0.750 \mathrm{~F}\) and you possess numerous \(1.50 \mathrm{mF}\) capacitors. What is the smallest number you could hook together to achieve your goal, and how would you connect them?

The electric field strength between two parallel conducting plates separated by \(4.00 \mathrm{~cm}\) is \(7.50 \times 10^{4} \mathrm{~V} / \mathrm{m}\). (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential \(1.00 \mathrm{~cm}\) from that plate (and \(3.00 \mathrm{~cm}\) from the other)?

Antiballistic missiles (ABMs) are designed to have very large accelerations so that they may intercept fast-moving incoming missiles in the short time available. What is the takeoff acceleration of a 10,000 -kg ABM that expels 196 kg of gas per second at an exhaust velocity of \(2.50 \times 10^{3} \mathrm{m} / \mathrm{s} ?\)

A 30,000 -kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of Scrap metal into it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost?

A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of \(4.00 \mathrm{m} / \mathrm{s}\). (a) What is the average force exerted on the leg, taking the effective mass of the hand and forearm to be \(1.50 \mathrm{kg}\) ? (b) Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not.

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