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A very small sphere with positive charge \(q=+8.00 \mu \mathrm{C}\) is released from rest at a point 1.50 \(\mathrm{cm}\) from a very long line of uniform linear charge density \(\lambda=+3.00 \mu \mathrm{Cm} .\) What is the kinetic energy of the sphere when it is 4.50 \(\mathrm{cm}\) from the line of charge if the only force on it is the force exerted by the line of charge?

Short Answer

Expert verified
The kinetic energy when the sphere is 4.50 cm from the line is due to the change in electric potential energy, which is converted into kinetic energy.

Step by step solution

01

Understand the Problem

We want to find the kinetic energy of a charged sphere after it has moved a certain distance in an electric field created by a line of charge. We start by noting the initial and final distances of the sphere from the line and that the sphere begins at rest.
02

Identify Key Formulas

The force on a charge in an electric field, created by a line of charge, is given by Coulomb's Law. However, since this is a line of charge, we use the electric field from a line, given by: \( E = \frac{\lambda}{2\pi\epsilon_0 r} \). The change in electric potential energy though, which relates to kinetic energy, uses the potential difference formula.
03

Calculate Initial and Final Electric Potential

The electric potential at a distance \( r \) from an infinite line of charge is \( V = -\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r}{r_0}\right) \). Calculate the potential at the initial point (1.5 cm) and the final point (4.5 cm).
04

Compute Change in Electric Potential Energy

Using the initial and final potentials, calculate the change in electric potential energy, \( \Delta U = q \cdot (V_{final} - V_{initial}) \).
05

Convert Potential Energy to Kinetic Energy

Since the sphere starts from rest, its initial kinetic energy is zero. Thus, the change in electric potential energy is equal to the kinetic energy gained, \( \Delta U = K \).
06

Solve the Calculation

Compute the values using \( \lambda = 3.00 \text{ }\mu\text{C/m}, \; q = 8.00 \text{ }\mu\text{C}, \; \epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2) \). Plug these values into the formulas, to find the final kinetic energy.

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Key Concepts

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

Electric Field
An electric field represents a region in space where electric charges experience a force. For a line of charge, the electric field behaves differently than for a point charge. Here, the electric field due to a very long line of charge with uniform linear charge density \lambda\ is given by the formula: \( E = \frac{\lambda}{2\pi\epsilon_0 r} \). This illustrates how the field strength decreases with increasing distance \( r \) from the line of charge.

\[ E \] denotes the electric field strength, \( \lambda \) is the linear charge density (charge per unit length), and \( \epsilon_0 \) is the vacuum permittivity. The electric field is crucial in determining how a charged particle, like the sphere in our problem, will move under the influence of forces.
Line of Charge
A line of charge refers to a configuration where charge is distributed uniformly along an infinitely long line. This differs from point charges, leading to unique electric field properties.

When dealing with a line of charge, the electric field at a point depend only on the perpendicular distance from the line rather than the distance from a single charge point. The formula \( E = \frac{\lambda}{2\pi\epsilon_0 r} \) highlights this relationship.

Such forms of charge distribution often simplify calculations and are key in understanding how electric fields can be applied in different scenarios. A practical example includes cables or charged rods that we encounter in real-world applications.
Coulomb's Law
Coulomb's Law describes the force between two charges. Essentially, it states that the force between static charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

While Coulomb’s Law is directly used for point charges, its principles help understand forces from lines of charge. The electric field termed earlier builds upon these basics to cater to line charges.

Understanding this law is essential as it forms the groundwork for evaluating electrical forces in many systems, allowing us to predict how charged particles will move under given conditions.
Electric Potential Energy
Electric potential energy is the energy a charge possesses due to its position in an electric field. In the case of a charged sphere moving in a field generated by a line of charge, potential energy changes as it moves closer or farther away.

The potential energy difference is calculated using \( \Delta U = q \cdot (V_{final} - V_{initial}) \) where \( q \) is the charge of the sphere, and \( V \) the potential at given points. Here, \( V \) in terms of a line of charge is given by \( V = -\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r}{r_0}\right) \).

This energy transformation is pivotal: when the sphere starts from rest, its initial kinetic energy is zero, so the entire change in potential energy converts to kinetic energy. Thus, knowing this difference gives us the kinetic energy of the sphere at its new position.

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Most popular questions from this chapter

A very long insulating cylindrical shell of radius 6.00 \(\mathrm{cm}\) carries charge of linear density 8.50\(\mu \mathrm{C} / \mathrm{m}\) spread uniformly over its outer surface. What would a voltmeter read if it were connected between (a) the surface of the cylinder and a point 4.00 \(\mathrm{cm}\) above the surface, and (b) the surface and a point 1.00 \(\mathrm{cm}\) from the central axis of the cylinder?

Deffecting Plates of an Oscilloscope. The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. Typical dimensions are about 3.0 \(\mathrm{cm}\) on a side, with a separation of about 5.0 \(\mathrm{mm} .\) The potential difference between the plates is 25.0 \(\mathrm{V} .\) The plates are close enough that we can ignore fringing at the ends. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 \(\mathrm{cm} .\) (a) If the surface charge density for each plate has magnitude 47.0 \(\mathrm{nC} / \mathrm{m}^{2}\) , what is the magnitude of \(\vec{\boldsymbol{E}}\) in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?

Two point charges are moving to the right along the \(x\) -axis. Point charge 1 has charge \(q_{1}=2.00 \mu \mathrm{C},\) mass \(m_{1}=\) \(6.00 \times 10^{-5} \mathrm{kg},\) and speed \(v_{1} .\) Point charge 2 is to the right of \(q_{1}\) and has charge \(q_{2}=-5.00 \mu \mathrm{C},\) mass \(m_{2}=3.00 \times 10^{-5} \mathrm{kg},\) and speed \(v_{2} .\) At a particular instant, the charges are separated by a distance of 9.00 \(\mathrm{mm}\) and have speeds \(v_{1}=400 \mathrm{m} / \mathrm{s}\) and \(v_{2}=1300 \mathrm{m} / \mathrm{s} .\) The only forces on the particles are the forces they exert on each other. (a) Determine the speed \(v_{\mathrm{cm}}\) of the center of mass of the system. (b) The relative energy \(E_{\text { rel }}\) of the system is defined as the total energy minus the kinetic energy contributed by the motion of the center of mass: $$E_{\mathrm{rel}}=E-\frac{1}{2}\left(m_{1}+m_{2}\right) v_{\mathrm{cm}}^{2}$$ where \(E=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r\) is the total energy of the system and \(r\) is the distance between the charges. Show that \(E_{\mathrm{rel}}=\frac{1}{2} \mu v^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r, \quad\) where \(\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)\) is called the reduced mass of the system and \(v=v_{2}-v_{1}\) is the relative speed of the moving particles. (c) For the numerical values given above, calculate the numerical value of \(E_{\text { rel. }}\) (d) Based on the result of part (c), for the conditions given above, will the particles escape from one another? Explain. (e) If the particles do escape, what will be their final relative speed when \(r \rightarrow \infty ?\) If the particles do not escape, what will be their distance of maximum separation? That is, what will be the value of \(r\) when \(v=0 ?\) (f) Repeat parts \((c)-(e)\) for \(v_{1}=400 \mathrm{m} / \mathrm{s}\) and \(v_{2}=1800 \mathrm{m} / \mathrm{s}\) when the separation is 9.00 \(\mathrm{mm} .\)

Self-Energy of a Sphere of Charge. A solid sphere of radius \(R\) contains a total charge \(Q\) distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the "self-energy" of the charge distribution. (Hint: After you have assembled a charge \(q\) in a sphere of radius \(r,\) how much energy would it take to add a spherical shell of thickness dr having charge \(d q ?\) Then integrate to get the total energy.)

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