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Chapter 23: Problem 30

A solid conducting sphere has net positive charge and radius \(R=0.400 \mathrm{~m}\). At a point \(1.20 \mathrm{~m}\) from the center of the sphere, the electric potential due to the charge on the sphere is \(24.0 \mathrm{~V}\). Assume that \(V=0\) at an infinite distance from the sphere. What is the electric potential at the center of the sphere?

Short Answer

Expert verified
The electric potential at the center of the sphere is 72.0 V.

Step by step solution

01

- Find the Charge (q)

Firstly, we use the given value for electric potential V=24.0 V at 1.20 m from the center of the sphere to solve for the charge q using equation \(q = \frac{Vr}{K}\). Substituting values to get q as \(q=\frac{(24.0V)(1.20m)}{(8.99 * 10^9 N m^2/C^2)} = 3.20 * 10^{-9} C\).
02

- Find the Potential at the Center

We then substitute the obtained value of q into the formula \(V=\frac{Kq}{r}\) to find the electric potential at the center of the sphere. But since r=R=0.4 m, we get \(V = \frac{(8.99 * 10^9 N m^2/C^2)(3.20 * 10^{-9} C)}{0.400m} = 72.0 V\).

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Key Concepts

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

Conducting Sphere Charge
When we deal with a conducting sphere that carries a charge, it's important to understand that the charges on a conductor in electrostatic equilibrium reside on the surface. This implies that within the conducting sphere, there is no excess charge. Hence, for practical purposes, the charge of the entire sphere behaves as if it's located at the center when calculating electric fields or potentials outside the conductor.

From the given problem, we use the concept of electric potential at a known distance and the principle that the potential due to a point charge (or a concentric spherical charge distribution) is given by the formula \(V = \frac{Kq}{r}\), where \(V\) is the electric potential, \(q\) is the charge, \(r\) is the distance from the charge to the point where the potential is being measured, and \(K\) is Coulomb's constant. Knowing the potential and distance, we can rearrange the formula to solve for the charge \(q\).
Electric Potential Formula
Understanding the electric potential formula is crucial for solving many problems in electrostatics. The formula, \(V = \frac{Kq}{r}\), can be viewed as a measure of the electric potential energy per unit charge at a specific point in an electric field, created by a charge \(q\).

For a point at the center of a charged conducting sphere, the potential is equivalent to the potential at any point on the surface. This is because the electric potential is constant throughout a conductor at equilibrium. Thus, to calculate this potential, we use the radius of the sphere as \(r\) in the formula, as all the surface charge theoretically acts from this radius, leading to the solution for the electric potential at the center.
Coulomb's Constant
Coulomb's constant (\(K\)), also known as the electric force constant, is an essential part of the electric potential formula. Its value is approximately \(8.99 \times 10^9 N m^2/C^2\), and it originates from Coulomb's law, which describes the force between two point charges. The constant relates the electric force between point charges to the product of the charges and inversely to the square of the distance between them.

The constant plays a fundamental role in calculating electric potentials as well. It determines how the charge's value and the distance between the charge influence the potential experienced at a point in the electric field generated by the charge. Thus, with the value of Coulomb's constant known, we can then predict the electric potential across a variety of configurations and distances.

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Most popular questions from this chapter

BIO Electrical Sensitivity of Sharks. Certain sharks can detect an electric field as weak as \(1.0 \mu \mathrm{V} / \mathrm{m}\). To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5 V AA battery across these plates, how far apart would the plates have to be?

CP Deflecting Plates of an Oscilloscope. The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying cqual 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 cnds. 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?

An annulus with an inner radius of \(a\) and an outer radius of \(b\) has charge density \(\sigma\) and lies in the \(x y\) -plane with its center at the origin, as shown in Fig. \(\mathbf{P} 2 \mathbf{3 . 8 0}\). (a) Using the convention that the potential vanishes at infinity, determine the potential at all points on the \(z\) -axis. (b) Determine the electric field at all points on the \(z\) -axis by differentiating the potential. (c) Show that in the limit \(a \rightarrow 0, b \rightarrow \infty\) the electric field reproduces the result obtained in Example 22.7 for an infinite plane sheet of charge. (d) If \(a=5.00 \mathrm{~cm}, b=10.0 \mathrm{~cm}\) and the total charge on the annulus is \(1.00 \mu \mathrm{C}\), what is the potential at the origin? (e) If a particle with mass \(1.00 \mathrm{~g}\) (much less than the mass of the annulus) and charge \(1.00 \mu \mathrm{C}\) is placed at the origin and given the slightest nudge, it will be projected along the \(z\) -axis. In this case, what will be its ultimate speed?

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?

A very long insulating cylinder of charge of radius \(2.50 \mathrm{~cm}\) carries a uniform linear density of \(15.0 \mathrm{nC} / \mathrm{m}\). If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads \(175 \mathrm{~V} ?\)
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