/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 The electric field at the surfac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The electric field at the surface of a charged, solid, copper sphere with radius \(0.200 \mathrm{~m}\) is \(3800 \mathrm{~N} / \mathrm{C}\), directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

Short Answer

Expert verified
The potential at the center of the sphere is -76.2 V.

Step by step solution

01

Identify given quantities

We have a solid, charged, copper sphere with a radius of \(0.200 \ m\) and an electric field at its surface of \(3800 \ N/C\), which is directed towards the sphere's center. The electric potential infinitively far from the sphere is taken to be zero.
02

Use Coulomb's Law to compute the charge

Coulomb's law for the electric field \(E\) of a charged sphere is given as \(E = k_e \frac{Q}{r^2}\) where \(k_e\) is the electrostatic constant, \(Q\) is the charge on the sphere, and \(r\) is the radius of the sphere. The electric field is directed towards the sphere, which indicates a negative charge. Solving for \(Q\), we find \(Q = E \cdot r^2 / k_e = 3800 \cdot (0.2)^2 / (8.99 \times 10^9) = -1.689 \times 10^{-9} \ C\).
03

Calculate the potential at the center

The electric potential \(V\) at a point located at distance \(r\) from a point charge \(Q\) is given by \(V = k_e \frac{Q}{r}\). Since we want to find the potential at the center of the sphere, \(r\) is equal to the radius of the sphere. Replacing \(r\) with the radius of the sphere and \(Q\) with the charge obtained earlier, we get \(V = k_e \frac{Q}{r} = 8.99 \times 10^9 \cdot (\frac{-1.689 \times 10^{-9}}{0.2}) = -76.2 \ V\)

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 91Ó°ÊÓ!

Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle that helps us understand how electric charges interact with each other. It describes the force between two stationary, point charges. According to Coulomb's Law, the electric force (F) between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This can be mathematically expressed as:\[F = k_e \frac{|Q_1Q_2|}{r^2}\]where:
  • \(F\) is the force between the charges,
  • \(Q_1\) and \(Q_2\) are the magnitudes of the charges,
  • \(r\) is the distance between the charges,
  • \(k_e\) is the electrostatic constant, approximately equal to \(8.99 \times 10^9 \, \text{N}·\text{m}^2/\text{C}^2\) .

In the context of a charged sphere, Coulomb's Law can be slightly modified since the sphere effectively behaves as a point charge at its center when calculating forces and fields at distances greater than or equal to its radius. This principle helps derive other properties, like the electric field, associated with the charged sphere.
Charged Sphere
A charged sphere, particularly when considered a solid metal sphere like copper, distributes charge uniformly across its surface. Inside a conductor, such as metal, the electric field is zero, and all excess charge resides on the surface.
The electric field (E) at the surface of a charged sphere can be determined using a form of Coulomb's Law. It follows:\[E = k_e \frac{Q}{r^2}\]Here:
  • \(E\) is the electric field on the surface,
  • \(Q\) is the total charge on the sphere,
  • \(r\) is the radius of the sphere.

In this exercise, the electric field was given as \(3800 \, \text{N/C}\), directed inward, indicating a negatively charged sphere. The surface's electric field tells us about how the sphere's charge interacts with the space around it. Since our example assumes potential to be zero at infinity, the calculated electric potential at any point can give insight into the energy required to move a charge to that point through the sphere’s electric field.
Electric Field
The electric field is a vector field around a charged object where each point in the space has a vector associated with it that points in the direction that a positive test charge would move if placed at that point. The magnitude of the electric field tells us the force that a charge of \(1 \, \text{C}\) would experience at each point.
The formula for the electric field \(E\) around a point charge \(Q\) is given as:\[E = k_e \frac{Q}{r^2}\]
  • \(E\) represents the electric field,
  • \(k_e\) is the electrostatic constant,
  • \(Q\) is the point charge,
  • \(r\) is the distance from the charge.

For a charged sphere, such an electric field will be strong and directed towards or away from the sphere depending on the charge’s sign. In this problem, since the field is toward the center, it suggests a negative surface charge. Understanding the electric field is essential for computing the behavior of charges around the sphere, and ultimately, determining the potential at different points within and around the sphere.

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

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are \(4.98 \mathrm{~V}\) and \(16.2 \mathrm{~V} / \mathrm{m}\) respectively. (Take \(V=0\) at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the noint charge?

In a certain region of space the electric potential is given by \(V=+A x^{2} y-B x y^{2},\) where \(A=5.00 \mathrm{~V} / \mathrm{m}^{3}\) and \(B=8.00 \mathrm{~V} / \mathrm{m}^{3}\) Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x=2.00 \mathrm{~m}, y=0.400 \mathrm{~m},\) and \(z=0\)

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) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to a halt?

A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a\), and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{~J}\). When the second charge is moved to point \(b,\) the clectric force on the charge does \(-1.9 \times 10^{-8} \mathrm{~J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b ?\)

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.