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 \(1436 \mathrm{V}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electron is accelerated from rest through a potential difference of \(370 \mathrm{~V}\). What is its final speed?

In molecules of gaseous sodium chloride, the chloride ion has one more electron than proton, and the sodium ion has one more proton than electron. These ions are separated by about \(0.24 \mathrm{nm}\). How much work would be required to increase the distance between these ions to \(1.0 \mathrm{~cm} ?\)

A thin line of charge is aligned along the positive \(y\) -axis from \(0 \leq y \leq L,\) with \(L=4.0 \mathrm{~cm} .\) The charge is not uniformly distributed but has a charge per unit length of \(\lambda=A y,\) with \(A=\) \(8.0 \cdot 10^{-7} \mathrm{C} / \mathrm{m}^{2}\). Assuming that the electric potential is zero at infinite distance, find the electric potential at a point on the \(x\) -axis as a function of \(x\). Give the value of the electric potential at \(x=3.0 \mathrm{~cm} .\)

A charge \(Q=\) \(+5.60 \mu C\) is uniformly distributed on a thin cylindrical plastic shell. The radius, \(R\), of the shell is \(4.50 \mathrm{~cm}\). Calculate the electric potential at the origin of the \(x y\) -coordinate system shown in the figure. Assume that the electric potential is zero at points infinitely far away from the origin.

A point charge \(Q\) is placed a distance \(R\) from the center of a conducting sphere of radius \(a,\) with \(R>a\) (the point charge is outside the sphere). The sphere is grounded, that is, connected to a distant, unlimited source and/or sink of charge at zero potential. (Neither the distant ground nor the connection directly affects the electric field in the vicinity of the charge and sphere.) As a result, the sphere acquires a charge opposite in sign to \(Q\), and the point charge experiences an attractive force toward the sphere. a) Remarkably, the electric field outside the sphere is the same as would be produced by the point charge \(Q\) plus an imaginary mirror-image point charge \(q\), with magnitude and location that make the set of points corresponding to the surface of the sphere an equipotential of potential zero. That is, the imaginary point charge produces the same field contribution outside the sphere as the actual surface charge on the sphere. Calculate the value and location of \(q\). (Hint: By symmetry, \(q\) must lie somewhere on the axis that passes through the center of the sphere and the location of \(Q .)\) b) Calculate the force exerted on point charge \(Q\) and directed toward the sphere, in terms of the original quantities \(Q, R,\) and \(a\) c) Determine the actual nonuniform surface charge distribution on the conducting sphere.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free