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Chapter 25: Problem 2

An ion accelerated through a potential difference of 115 \(\mathrm{V}\) experiences an increase in kinetic energy of \(7.37 \times 10^{-17} \mathrm{J}\) . Calculate the charge on the ion.

Short Answer

Expert verified
The charge on the ion is \(6.41 \times 10^{-19} \textrm{C}\).

Step by step solution

01

Understanding the relationship between voltage, charge, and kinetic energy

The work done on the charge by the electric field as it moves through a potential difference (voltage) results in a change in its kinetic energy. Mathematically, this is represented by the equation: \( W = qV \), where \( W \) is the work done or the change in kinetic energy, \( q \) is the charge of the ion, and \( V \) is the potential difference.
02

Rearrange the equation to solve for the charge

To find the charge on the ion, we need to rearrange the equation to solve for \( q \): \( q = \frac{W}{V} \).
03

Substitute the known values into the equation

Substitute the kinetic energy \( 7.37 \times 10^{-17} \textrm{J} \) for \( W \) and \( 115 \textrm{V} \) for \( V \) in the equation to find \( q \): \( q = \frac{7.37 \times 10^{-17} \textrm{J}}{115 \textrm{V}} \).
04

Calculate the charge

Perform the division to get the value of the charge \( q \): \( q = \frac{7.37 \times 10^{-17} \textrm{J}}{115 \textrm{V}} = 6.41 \times 10^{-19} \textrm{C} \). Note that the unit of charge is coulombs (\( \text{C} \)).

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Key Concepts

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

Electric Potential Difference
When we talk about electric potential difference, often referred to as voltage, we are essentially discussing the 'push' that causes charges to move. Think of it as the difference in electric potential energy per unit charge between two points in an electric field. The larger the difference, the greater the propensity for charges to move from one point to another.

As seen in our exercise, an ion moving across a difference of 115 volts is gaining energy due to this electric push. For those not familiar, a 'volt' is the derived unit for electric potential, electric potential difference, and electromotive force. Essentially, each volt corresponds to one joule of energy per coulomb of charge that moves. This relationship is crucial as it underpins the way electric fields do work on charges, tying directly into the equation we used: \( W = qV \).
Kinetic Energy
Kinetic energy is the energy of motion. Every moving object has kinetic energy, and it鈥檚 proportional to the mass of the object and the square of its velocity. The formula we use to describe this relationship is: \( KE = \frac{1}{2}mv^2 \), where \( KE \) is the kinetic energy, \( m \) is the mass, and \( v \) is the velocity of the object.

In the context of charged particles, such as the ion in our exercise, when they are accelerated through an electric field by a potential difference, this energy transition is from electric potential energy to kinetic energy. This is visible in the exercise as the result of an ion gaining a kinetic energy of \( 7.37 \times 10^{-17} \, \mathrm{J} \) upon being accelerated.
Work-Energy Principle in Physics
The work-energy principle is a cornerstone of physics that states that work done on an object is equal to the change in its kinetic energy. The formula for work done is: \( W = Fd \), where \( W \) is the work, \( F \) is the force and \( d \) is the displacement in the direction of the force. When related to electric fields and potential difference, the work done on a charge by the electric force is manifest as the charge's kinetic energy change.

Our solution applies this principle, using the fact that the work done by the electric field on the ion is equal to the product of the ion's charge and the electric potential difference it travels through. This translates to the formula \( W = qV \) 鈥� a relation that simplifies our calculations and directly connects the concepts of work, energy, voltage, and charge.

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Most popular questions from this chapter

(a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 \(\mathrm{V}\) . (b) Calculate the speed of an electron that is accelerated through the same potential difference.

Electric charge can accumulate on an airplane in flight. You may have observed needle-shaped metal extensions on the wing tips and tail of an airplane. Their purpose is to allow charge to leak off before much of it accumulates. The electric field around the needle is much larger than the field around the body of the airplane, and can become large enough to produce dielectric breakdown of the air, discharging the airplane. To model this process, assume that two charged spherical conductors are connected by a long conducting wire, and a charge of 1.20\(\mu \mathrm{C}\) is placed on the combination. One sphere, representing the body of the airplane, has a radius of \(6.00 \mathrm{cm},\) and the other, representing the tip of the needle, has a radius of \(2.00 \mathrm{cm} .\) (a) What is the electric potential of each sphere? (b) What is the electric field at the surface of each sphere?

Review problem. Two insulating spheres have radii 0.300 \(\mathrm{cm}\) and \(0.500 \mathrm{cm},\) masses 0.100 \(\mathrm{kg}\) and \(0.700 \mathrm{kg},\) and uniformly distributed charges of \(-2.00 \mu \mathrm{C}\) and 3.00\(\mu \mathrm{C}\) . They are released from rest when their centers are separated by 1.00 \(\mathrm{m}\) (a) How fast will each be moving when they collide? (Suggestion: consider conservation of energy and of linear momentum.) (b) What If? If the spheres were conductors, would the speeds be greater or less than those calculated in part (a)? Explain.

A solid sphere of radius \(R\) has a uniform charge density \(\rho\) and total charge \(Q\) . Derive an expression for its total electric potential energy. (Suggestion: imagine that the sphere is constructed by adding successive layers of concentric shells of charge \(d q=\left(4 \pi r^{2} d r\right) \rho\) and use \(d U=V d q\) .

Consider a ring of radius \(R\) with the total charge \(Q\) spread uniformly over its perimeter. What is the potential difference between the point at the center of the ring and a point on its axis a distance 2\(R\) from the center?
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