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Chapter 23: Problem 34

A hollow spherical conductor with a \(5.0-\mathrm{cm}\) radius has a surface charge of \(8.0 \mathrm{nC}\). a) What is the potential \(8.0 \mathrm{~cm}\) from the center of the sphere? b) What is the potential \(3.0 \mathrm{~cm}\) from the center of the sphere? c) What is the potential at the center of the sphere?

Short Answer

Expert verified
Answer: The electric potentials at these distances are as follows: - At 8.0 cm from the center of the sphere: 898 V - At 3.0 cm from the center of the sphere: 1436 V - At the center of the sphere: 1436 V

Step by step solution

01

Calculate distance from the sphere's surface

First, we need to find the distance from the surface of the sphere to the point where we want to determine the potential, which is 8.0 cm from the center of the sphere. Since the radius is 5.0 cm, this point is \((8.0 - 5.0) = 3.0 \mathrm{cm}\) away from the sphere's surface.
02

Apply the electric potential formula

Now we will use the formula \(V = \frac{kQ}{r}\) to calculate the potential. We convert the given charge from nC to C by multiplying with \(10^{-9}\) and the distance from cm to meters by multiplying with \(10^{-2}\). $$V = \frac{8.99 \times 10^9 \mathrm{Nm^2/C^2} \times 8.0 \times 10^{-9} \mathrm{C}}{(8.0 \times 10^{-2}) \mathrm{m}} = 898 \mathrm{V}$$ The potential at the point \(8.0 \mathrm{~cm}\) from the center of the sphere is \(898 \mathrm{V}\). #b) Potential 3.0 cm from the center of the sphere#
03

Constant potential inside the sphere

Since it is a hollow conductor, the potential inside the sphere is constant and equal to the potential at the surface.
04

Apply the electric potential formula

We will use the formula \(V = \frac{kQ}{r}\) to calculate the potential at the surface of the sphere: $$V = \frac{8.99 \times 10^9 \mathrm{Nm^2/C^2} \times 8.0 \times 10^{-9} \mathrm{C}}{(5.0 \times 10^{-2}) \mathrm{m}} = 1436 \mathrm{V}$$ The potential at the point \(3.0 \mathrm{~cm}\) from the center of the sphere is \(1436 \mathrm{V}\). #c) Potential at the center of the sphere#
05

Constant potential inside the sphere

As mentioned previously, the potential inside the sphere is constant and equal to the potential at the surface. The potential at the center of the sphere is \(1436 \mathrm{V}\).

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Most popular questions from this chapter

Why is it important, when soldering connectors onto a piece of electronic circuitry, to leave no pointy protrusions from the solder joints?

An electric field is established in a nonuniform rod. A voltmeter is used to measure the potential difference between the left end of the rod and a point a distance \(x\) from the left end. The process is repeated, and it is found that the data are described by the relationship \(\Delta V=270 x^{2},\) where \(\Delta V\) has the units \(\mathrm{V} / \mathrm{m}^{2}\). What is the \(x\) -component of the electric field at a point \(13 \mathrm{~cm}\) from the left end?

Two fixed point charges are on the \(x\) -axis. A charge of \(-3.00 \mathrm{mC}\) is located at \(x=+2.00 \mathrm{~m}\) and a charge of \(+5.00 \mathrm{mC}\) is located at \(x=-4.00 \mathrm{~m}\) a) Find the electric potential, \(V(x),\) for an arbitrary point on the \(x\) -axis. b) At what position(s) on the \(x\) -axis is \(V(x)=0 ?\) c) Find \(E(x)\) for an arbitrary point on the \(x\) -axis.

A proton is placed midway between points \(A\) and \(B\). The potential at point \(A\) is \(-20 \mathrm{~V}\), and the potential at point \(B\) \(+20 \mathrm{~V}\). The potential at the midpoint is \(0 \mathrm{~V}\). The proton will a) remain at rest. b) move toward point \(B\) with constant velocity. c) accelerate toward point \(A\). d) accelerate toward point \(B\). e) move toward point \(A\) with constant velocity.

Each of the following pairs of charges are separated by a distance \(d\). Which pair has the highest potential energy? a) \(+5 \mathrm{C}\) and \(+3 \mathrm{C}\) d) \(-5 \mathrm{C}\) and \(+3 \mathrm{C}\) b) \(+5 \mathrm{C}\) and \(-3 \mathrm{C}\) e) All pairs have the \(\begin{array}{ll}\text { c) }-5 \mathrm{C} \text { and }-3 \mathrm{C} & \text { same potential energy. }\end{array}\)
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