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Chapter 23: Problem 50

A small sphere with mass \(5.00 \times 10^{-7} \mathrm{~kg}\) and charge \(+7.00 \mu \mathrm{C}\) is released from rest a distance of \(0.400 \mathrm{~m}\) above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma=+8.00 \mathrm{pC} / \mathrm{m}^{2} .\) Using energy methods, calculate the speed of the sphere when it is \(0.100 \mathrm{~m}\) above the sheet.

Short Answer

Expert verified
The speed of the sphere when it is 0.100m above the sheet is \(1.38 m/s\)

Step by step solution

01

Calculate the initial potential energy of the system

We calculate the initial potential energy of the system when the sphere is at a distance 0.400m from the sheet. The formula for the potential energy \(U\) due to a uniformly charged infinite sheet is given by, \(U = \frac{1}{2}qV\), where \(q\) is the charge and \(V\) is the potential. The electric potential \(V\) of a uniformly charged infinite sheet is given by \(V= \frac{\sigma d}{\varepsilon_0}\), where \(\sigma\) is the charge density, \(d\) is the perpendicular distance and \(\varepsilon_0\) is the permittivity of free space. Thus, substituting the value of \(V\) we get, \(U = \frac{q\sigma d}{2\varepsilon_0}\). Substituting the known values, we get \(U = \frac{(7.00 \times 10^{-6})(8.00 \times 10^{-12})(0.400)}{2(8.85 \times 10^{-12})}\) resulting in \(U = 1.26 \times 10^{-6}J\).
02

Calculate the final potential energy

Next, we calculate the final potential energy, when the sphere is 0.100m from the sheet. Using the formula from previous step, we substitute \(d = 0.100m\) and we get final potential energy \(U' = \frac{(7.00 \times 10^{-6})(8.00 \times 10^{-12})(0.100)}{2(8.85 \times 10^{-12})}\) resulting in \(U' = 3.15 \times 10^{-7}J\).
03

Calculate the kinetic energy and the speed of the sphere

The kinetic energy is the difference between the initial potential energy and the final potential energy, \(K = U - U'\). Substitute the values we get, \(K = 1.26 \times 10^{-6} - 3.15 \times 10^{-7} = 9.45 \times 10^{-7}J\). The kinetic energy is also given by \(K = \frac{1}{2}mv^{2}\) where \(m\) is the mass and \(v\) is the speed. Substituting \(K\) and \(m = 5.00 \times 10^{-7}kg\), we can solve for \(v\). Thus, \(5.00 \times 10^{-7}v^{2} = 9.45 \times 10^{-7}\). Solving it, we can find \(v = 1.38 m/s\).

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Key Concepts

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

Uniform Surface Charge Density
When dealing with electric charges distributed over a surface, the concept of uniform surface charge density is key. If a large sheet has a uniform distribution of charge, it means the charge is spread evenly over its surface. This helps in simplifying calculations involving electric fields and electric potentials.

The surface charge density, usually denoted as \(\sigma\), is the amount of charge per unit area on the sheet. For a uniformly charged infinite sheet, it can impact a nearby charge particle's motion significantly.
  • It is measured in units such as coulombs per square meter (\(\mathrm{C/m}^{2}\)).
  • In our problem, the surface charge density of \(8.00 \mathrm{pC/m}^{2}\) means that each square meter of the sheet holds a charge of \(8.00\) picocoulombs.
This concept helps in calculating electric potential, which in turn affects how the potential energy of a charge changes with distance.
Electrostatics
Electrostatics is the branch of physics that studies electric charges at rest. In this context, it explains how charged objects interact with each other and how an electric field results from a charge distribution like that of an insulating sheet with a uniform surface charge density.

In electrostatics, the electric field generated by a uniformly charged infinite sheet is continuous and uniform. This is a simplifying assumption helpful in solving many problems, as it means the electric field does not change with distance from the surface.
  • This uniform field simplifies calculations of electric potential.
  • The potential energy of a charged particle in this field depends on its position relative to the sheet.
  • Electric potential \(V\) from such a sheet is given by \(V = \frac{\sigma d}{\varepsilon_0}\), which depends linearly on the distance \(d\) from the sheet.
Understanding these principles is crucial when calculating changes in potential energy, as seen in the provided problem.
Kinetic Energy
Kinetic energy is the energy of motion, and it plays a critical role when considering the movement of a charged particle under the influence of an electric field.

When a charge is released in an electric field, like in the problem, it gains kinetic energy as it accelerates towards an opposite charge or away from a like charge. The change in potential energy of a charge translates into kinetic energy. This energy conversion is governed by the conservation of energy principle:
  • The total energy (potential plus kinetic) remains constant in a closed system.
  • As the particle moves closer to the sheet, its potential energy decreases and this lost energy appears as an increased kinetic energy.
  • In the example, the calculation of kinetic energy shows how much speed the sphere gains from a change in potential energy as it descends.
These energy principles allow us to determine the speed of the sphere as it approaches the sheet, which is found to be \(1.38 \mathrm{m/s}\) in the given problem.

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Most popular questions from this chapter

Identical charges \(q=+5.00 \mu \mathrm{C}\) are placed at opposite corners of a square that has sides of length \(8.00 \mathrm{~cm}\). Point \(A\) is at one of the empty corners, and point \(B\) is at the center of the square. A charge \(q_{0}=-3.00 \mu \mathrm{C}\) is placed at point \(A\) and moves along the diagonal of the square to point \(B\). (a) What is the magnitude of the net electric force on \(\bar{q}_{0}\) when it is at point \(A ?\) Sketch the placement of the charges and the direction of the net force. (b) What is the magnitude of the net electric force on \(q_{0}\) when it is at point \(B ?\) (c) How much work does the electric force do on \(q_{0}\) during its motion from \(A\) to \(B ?\) Is this work positive or negative? When it goes from \(A\) to \(B,\) does \(q_{0}\) move to higher potential or to lower potential?

A particle with charge \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed to the left. The charge is released from rest and moves to the left; after it has moved \(6.00 \mathrm{~cm},\) its kinetic energy is \(+2.20 \times 10^{-6} \mathrm{~J}\). What are (a) the work done by the electric force, (b) the potential of the starting point with respect to the end point, and (c) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) ?

The electric potential \(V\) in a region of space is given by $$ V(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right) $$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\vec{E}\) at any point in this region. (b) The work done by the field when a \(1.50 \mu \mathrm{C}\) test charge moves from the point \((x, y, z)=(0,0,0.250 \mathrm{~m})\) to the origin is measured to be \(6.00 \times 10^{-5}\) J. Determine \(A\). (c) Determine the electric field at the point \((0,0,0.250 \mathrm{~m})\). (d) Show that in every plane parallel to the \(x z\) -plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{~V}\) and \(y=2.00 \mathrm{~m} ?\)

Two large, parallel, metal 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 \(+2.40 \mathrm{nC} ?\) (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

A helium ion (He \(^{+7}\) ) 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? (I \(\mathrm{fm}=1 \times 10^{-15} \mathrm{~m},\) and \(e\) is the magnitude of the charge of an electron or a proton. \((\mathrm{a}) 2 e ;(\mathrm{b}) 11 e ;(\mathrm{c}) 20 \mathrm{e} ;(\mathrm{d}) 22 e\)
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