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Chapter 23: Problem 52

A small sphere with mass \(5.00 \times 10^{-7} \mathrm{kg}\) and charge \(+3.00 \mu C\) 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 of charge?

Short Answer

Expert verified
The speed of the sphere is approximately 7.17 m/s when it is 0.100 m above the sheet.

Step by step solution

01

Understand the Problem

We have a small sphere with a given mass and charge that starts from rest and falls toward a charged sheet. We need to calculate its speed at a certain distance above the sheet using energy conservation principles.
02

Identify Relevant Formulas

The energy conservation principle involves equating the change in electric potential energy to the change in kinetic energy. The formula for electric potential energy change \[ \Delta U = q \cdot \Delta V \]where \( q \) is the charge of the sphere and \( \Delta V \) is the change in electric potential. The kinetic energy at a height \( h \):\[ K = \frac{1}{2} m v^2 \].
03

Calculate Electric Field from Surface Charge Density

The electric field \( E \) due to a large sheet with surface charge density \( \sigma \) is given by \[ E = \frac{\sigma}{2\varepsilon_0} \]where \( \varepsilon_0 \) is the permittivity of free space \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2} \). Substitute \( \sigma = 8.00 \, \mathrm{pC/m^2} \) into the formula to find \( E \).
04

Calculate Change in Electric Potential

To calculate the change in electric potential (\( \Delta V \)), use \[ \Delta V = E \cdot \Delta h \]where \( \Delta h \) is the change in height from \( 0.400 \, \mathrm{m} \) to \( 0.100 \, \mathrm{m} \).
05

Calculate Change in Electric Potential Energy

Use \[ \Delta U = q \cdot \Delta V \],with \( q = 3.00 \times 10^{-6} \, \mathrm{C} \) to find the change in electric potential energy as the sphere moves closer to the sheet.
06

Apply Energy Conservation

The total mechanical energy is conserved, implying that the initial potential energy change is converted into kinetic energy. Therefore, solve \[ \Delta U = \frac{1}{2} m v^2 \]for \( v \), where \( m = 5.00 \times 10^{-7} \, \mathrm{kg} \).
07

Solve for the Speed

Rearrange and solve the equation \[ v = \sqrt{\frac{2 \times \Delta U}{m}} \]to find the speed \( v \) of the sphere when it is \( 0.100 \, \mathrm{m} \) above the charged sheet.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. In the context of our small sphere, we are interested in determining its kinetic energy as it approaches the charged sheet from an initial state of rest. The formula for calculating kinetic energy, denoted as \( K \), is given by:

\[ K = \frac{1}{2} m v^2 \]
where:
  • \( m \) is the mass of the object (in this case, the sphere).
  • \( v \) is the speed of the object.
The sphere starts with zero kinetic energy (since it is at rest) and gains kinetic energy as it falls toward the charged sheet. By the time the sphere is 0.100 meters above the sheet, its kinetic energy is maximized due to the conversion of potential energy into kinetic energy. Understanding kinetic energy helps us calculate the velocity of the sphere by rearranging the formula to solve for \( v \), which is critical for finding the speed of the sphere at the desired height.
Electric Field
An electric field is a region around a charged object where force is exerted on other charged objects. In this problem, the electric field is created by the charged sheet, which has a uniform surface charge density \( \sigma \). The formula to find the magnitude of the electric field \( E \) due to a large sheet is:

\[ E = \frac{\sigma}{2\varepsilon_0} \]
where:
  • \( \sigma \) is the surface charge density (given as \( 8.00 \, \mathrm{pC/m^2} \)).
  • \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2} \)).
The electric field indicates how strong the influence of the charged sheet is over a certain distance. For this sphere, moving towards the sheet implies experiencing a force due to this field, thereby affecting its motion and energy states. This concept is vital in calculating the electric potential energy changes experienced by the sphere.
Energy Conservation
Energy conservation is a fundamental principle stating that within a closed system, total energy remains constant. It involves converting one form of energy into another without any loss. In the exercise with the sphere, we apply energy conservation by equating the change in electric potential energy to the change in kinetic energy, ensuring the total mechanical energy remains the same throughout the sphere's movement.

The change in electric potential energy \( \Delta U \) is calculated using:
\[ \Delta U = q \cdot \Delta V \]
where:
  • \( q \) is the charge of the sphere (\( 3.00 \mu C \)).
  • \( \Delta V \) is the change in electric potential.
As the sphere moves closer to the sheet, its potential energy decreases, converting into kinetic energy. By calculating these changes and utilizing the formula:
\[ \frac{1}{2} m v^2 = \Delta U \]
we find the speed of the sphere. This principle helps ensure accurate computations and predictions of the sphere's dynamics as it interacts with the electric field of the charged sheet.

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Most popular questions from this chapter

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{cm} . \mathrm{A}\) small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 \(\mathrm{cm}\) of the surface of the large sphere?

In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density \(\rho(r)\) depends on the distance \(r\) from the center of the distribution but not on the spherical polar angles \(\theta\) and \(\phi .\) The electric potential \(V(r)\) due to this charge distribution is $$V(r)=\left\\{\begin{array}{l}{\frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a} \\ {0} & {\text { for } r \geq a}\end{array}\right.$$ where \(\rho_{0}\) is a constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) and \(a\) is a constant having units of meters. (a) Derive expressions for \(\vec{E}\) for the regions \(r \leq a\) and \(r \geq a .[\)Hint\(:\) Use Eq. \((23.23) .]\) Explain why \(\vec{\boldsymbol{E}}\) has only a radial component. (b) Derive an expression for \(\rho(r)\) in each of the two regions \(r \leq a\) and \(r \geq a .[\)Hint t: Use Gauss's law for two spherical shells, one of radius \(r\) and the other of radius \(r+d r .\) The charge contained in the infinitesimal spherical shell of radius \(d r\) is \(d q=4 \pi r^{2} \rho(r) d r . ](c)\) Show that the net charge contained in the volume of a sphere of radius greater than or equal to \(a\) is zero. [Hint: Integrate the expressions derived in part (b) for \(\rho(r)\) over a spherical volume of radius greater than or equal to \(a.\)) Is this result consistent with the electric field for \(r>a\) that you calculated in part (a)?

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is not a linear function of the position, even with planar geometry, but is given by $$V(x)=C x^{4 / 3}$$ where \(x\) is the distance from the cathode and \(C\) is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 \(\mathrm{mm}\) and the potential difference between electrodes is 240 \(\mathrm{V}\) . (a) Determine the value of \(C .\) (b) Obtain a formula for the electric field between the electrodes as a function of \(x\) . (c) Determine the force on an electron when the electron is halfway between the electrodes.

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 \(\mathrm{mm} .\)

A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. A thin wire lies on the axis of a hollow metal cylinder and is insulated from it (Fig. P23.64). A large potential difference is established between the wire and the outer cylinder, with the wire at higher potential; this sets up a strong electric field directed radially out- ward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced are accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audible "click." Suppose the radius of the central wire is 145\(\mu \mathrm{m}\) and the radius of the hollow cylinder is 1.80 \(\mathrm{cm} .\) What potential difference between the wire and the cylinder produces an electric field of \(2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\) at a distance of 1.20 \(\mathrm{cm}\) from the axis of the wire? (The wire and cylinder are both very long in comparison to their radii, so the results of Problem 23.63 apply.)
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