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Chapter 3: Problem 27

A very large plastic sheet carries a uniform charge density of \(-6.00 \mathrm{nC} / \mathrm{m}^{2}\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by \(1.00 \mathrm{~V}\). What type of surfaces are these?

Short Answer

Expert verified
(a) The potential decreases; no reference needed. (b) Surface spacing: 0.00295 m. They are parallel planes.

Step by step solution

01

Understanding Electric Potential

The electric potential decreases as you move away from a negatively charged sheet. This is because a negative charge creates an electric field that points towards the sheet. As you move against the field direction, potential decreases. The potential difference does not depend on the reference point as it involves changes rather than absolute values.
02

Electric Field of an Infinite Charged Sheet

The electric field due to a uniformly charged infinite sheet with charge density \( \sigma \), where \( \sigma = -6.00 \, \mathrm{nC/m^2} \), is uniform and given by \( E = \frac{\sigma}{2\varepsilon_0} \). For \( \sigma = -6.00 \, \mathrm{nC/m^2} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2/(N \cdot m^2)} \), we can calculate \( E \).
03

Calculate Electric Field

Substituting the values, \( E = \frac{-6.00 \times 10^{-9} \, \mathrm{C/m^2}}{2 \cdot 8.85 \times 10^{-12} \, \mathrm{C^2/(N \cdot m^2)}} = -339.44 \, \mathrm{N/C} \). The negative sign indicates the field direction towards the sheet.
04

Equipotential Surface Spacing

The potential difference \( \Delta V \) between equipotential surfaces is given by \( \Delta V = E \cdot d \), where \( d \) is the separation. To find \( d \) for \( \Delta V = 1 \, \mathrm{V} \), \( d = \frac{1}{339.44} = 0.00295 \, \mathrm{m} \).
05

Type of Surfaces

These equipotential surfaces are planes parallel to the charged sheet, since the electric field is uniform and perpendicular to the surface.

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Key Concepts

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

Equipotential Surfaces
Equipotential surfaces are like economic levels on a topographic map, representing regions where the electric potential is the same. In this scenario with the large plastic sheet, these surfaces form planes that are parallel to the sheet. This occurs because the electric field is constant and uniform, pointing towards the negatively charged surface.

Moving along any equipotential surface requires no work because there is no change in electric potential. This is similar to walking along a path of the same height on a hill 鈥� your altitude, and thus potential energy, remains unchanged. For our charged sheet, the planes of constant potential help us understand how electric forces are distributed relative to the sheet. They illustrate areas where a charge would not naturally move unless acted on by another force.

Remember, these planes are spaced evenly, and for the potential to drop by 1 volt, the surfaces are approximately 0.00295 meters apart, determined by the uniform electric field associated with the charge density.
Electric Field
An electric field is a region around a charged object where the force is exerted on other charges. For an infinite charged sheet, like the one in the problem, the electric field is special because it remains consistent in strength and direction perpendicular to the sheet.

The formula for the electric field from an infinitely charged sheet is given as:\[E = \frac{\sigma}{2\varepsilon_0}\]where \(\sigma\) is the charge density, and \(\varepsilon_0\) is the permittivity of free space. The calculated value of \(E\) in this instance is \(-339.44 \, \mathrm{N/C}\). The negative sign indicates that the electric field directions are towards the sheet, aligning with the direction of an inward force on a positive test charge.

In uniform fields like this, the concept of field lines can help visualize how potential changes. Moving against these lines (or away from the sheet) decreases the potential, reflecting the work needed to move a charge against the force. Thus, understanding the uniform electric field helps predict and explain how potential and forces interact near the charged surface.
Charge Density
Charge density represents the amount of electric charge per unit area on the surface of an object. It is a critical parameter in determining how electric fields and potentials behave around the charged object. In our problem, a charge density \(-6.00 \, \mathrm{nC/m^2}\) is applied to the plastic sheet.

This uniform charge density impacts the electric field which remains constant and uniform everywhere. **Why does this matter?** Because knowing the charge density and its uniformity allows us to calculate consistently what occurs in the space surrounding the charged object, providing a clearer picture of the electric potential landscape.

With a higher charge density, you'd observe a stronger electric field and thus a smaller spacing between equipotential surfaces for a given change in potential. Consequently, charge density is a foundational concept in electrostatics crucial for understanding interactions and forces in electric contexts.

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Most popular questions from this chapter

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 $$ V(r)=\left\\{\begin{array}{l} \frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a \\ 0 \quad \text { for } r \geq a \end{array}\right. $$ 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 \(\overrightarrow{\boldsymbol{E}}\) for the regions \(r \leq a\) and \(r \geq\) a. [Hint: Use Eq. (23.23).] Explain why \(\vec{E}\) 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: 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 \(\left.d q=4 \pi r^{2} \rho(r) d r .\right]\) (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)?

Charges \(+q,-q,+q,-q\) areplaced at corners \(A B C D\) of a square of a side \(a\). A charge \(+q\) is placed at its centre. Find the interaction energy of the system

* A metal sphere with radius \(R_{1}\) has a charge \(Q_{1} .\) Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

* Charge \(Q=+4.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=5.00 \mathrm{~cm}\). What is the potential difference between the center of the sphere and the surface of the sphere?

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. Obtain a formula for the electric field between the electrodes as a function of \(x\)
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