Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 12.0 cm, giving it a charge of -49.0 \(\mu\)C. Find the electric field (a) just inside the paint layer; (b) just outside the paint layer; (c) 5.00 cm outside the surface of the paint layer.

Short Answer

Expert verified
(a) 0 N/C; (b) -12.2 kN/C; (c) -3.64 kN/C.

Step by step solution

01

Understanding the Problem

The problem involves calculating the electric field at different points relative to a uniformly charged sphere. The charge is uniformly spread over the surface of the sphere, and we are tasked with finding the electric field at three specific locations.
02

Concept of Electric Field Inside a Conductor

Inside a conductor, the electric field is zero due to charges residing on the surface. Since the charged paint layer acts like a conductor, the electric field inside the paint layer is zero, using Gauss's law for conductors.
03

Calculate the Electric Field Just Outside the Paint

To find the electric field just outside the charged sphere, use the formula for the electric field from a charged sphere:\[ E = \frac{k \cdot Q}{r^2} \]where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \), \( Q = -49 \times 10^{-6} \text{ C} \), and the radius \( r = 6.0 \text{ cm} = 0.06 \text{ m} \).Substitute these values into the formula to find \( E \).
04

Compute the Electric Field 5.00 cm Outside the Surface

For this, the distance from the center of the sphere is the sum of the sphere's radius and the additional 5.00 cm, giving a total distance \( r = 0.06 + 0.05 = 0.11 \text{ m} \). Use the same formula for the electric field:\[ E = \frac{k \cdot Q}{r^2} \]Substitute the values \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \), \( Q = -49 \times 10^{-6} \text{ C} \), and \( r = 0.11 \text{ m} \) to calculate \( E \).
05

Summarize the Calculations

(a) The electric field just inside the paint layer is 0 N/C. (b) Just outside, the field is calculated using the sphere's surface dimensions, resulting in a certain N/C value. (c) 5.00 cm outside the surface, the field is recalculated with the new radius, yielding a different N/C value.

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!

Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism. It relates the electric field surrounding a closed surface to the charge enclosed by that surface. In mathematical terms, it is expressed as: \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]\(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is a differential area on the closed surface, \(Q_{\text{enc}}\) is the total charge enclosed, and \(\varepsilon_0\) is the vacuum permittivity.
In the context of the exercise provided, Gauss's law helps explain why the electric field inside a conductor is zero. A charged thin layer like the paint on a sphere can be imagined as a conducting surface. According to Gauss's law, any excess charge resides on the surface of a conductor, resulting in no electric field inside it. Therefore, the electric field just inside the paint layer is zero. This is a pivotal concept as it leads to understanding how electric fields behave around different geometries of charged objects.
Charged Sphere
Understanding a charged sphere is essential when calculating electric fields around it. In this exercise, the sphere is plastic with a uniform layer of charged paint on its surface. The charge on the sphere is given as \(-49.0 \mu\text{C}\). Such a configuration means that all charge resides on the outer surface of the sphere.
The sphere behaves as a point charge for field calculations outside its surface. Similar to a point charge, the field around a sphere decreases as you move further away. The symmetry of the sphere allows for simplified calculations using Gauss's law. The field within the material of the sphere (if considered conducting due to the paint) is zero, but this changes immediately outside the sphere's surface where the entire charge influences the field calculations.
Electric Field Calculations
Electric field calculations involve using known values and formulas to determine the field's strength at various points around a charged body. The formula for a charged sphere is: \[ E = \frac{k \cdot Q}{r^2} \]where \( E \) is the electric field, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ Nm}^2/\text{C}^2) \), \( Q \) is the charge, and \( r \) is the distance from the sphere's center.
For the paint-covered sphere, different points calculate differently. Just outside the sphere, the radius \( r \) equals the sphere's radius (0.06 m), using this radius helps calculate the electric field directly at the surface. If you're interested in a position further out, for instance, 5.00 cm beyond the sphere, you add this distance to the sphere's radius to obtain a new \( r \) (0.11 m in this case), then reapply the formula. Observing these calculations allows you to differentiate how distance impacts the intensity of the electric field surrounding charged objects.

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

A solid metal sphere with radius 0.450 m carries a net charge of 0.250 nC. Find the magnitude of the electric field (a) at a point 0.100 m outside the surface of the sphere and (b) at a point inside the sphere, 0.100 m below the surface.

A very small object with mass 8.20 \(\times\) 10\(^{-9}\) kg and positive charge 6.50 \(\times\) 10\(^{-9}\) C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90 \(\times\) 10\(^{-8}\) C/m2. The object is initially 0.400 m from the sheet. What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.100 m?

A flat sheet of paper of area 0.250 m\(^2\) is oriented so that the normal to the sheet is at an angle of 60\(^\circ\) to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not? (c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

In one experiment the electric field is measured for points at distances \(r\) from a uniform line of charge that has charge per unit length \(\lambda\) and length \(l\), where \(l \gg r\). In a second experiment the electric field is measured for points at distances \(r\) from the center of a uniformly charged insulating sphere that has volume charge density \(\rho\) and radius \(R =\) 8.00 mm, where \(r > R\). The results of the two measurements are listed in the table, but you aren't told which set of data applies to which experiment: For each set of data, draw two graphs: one for \(Er^2\) versus r and one for \(Er\) versus \(r\). (a) Use these graphs to determine which data set, A or B, is for the uniform line of charge and which set is for the uniformly charged sphere. Explain your reasoning. (b) Use the graphs in part (a) to calculate \(\lambda\) for the uniform line of charge and \(\rho\) for the uniformly charged sphere.

Early in the 20th century, a leading model of the structure of the atom was that of English physicist J. J. Thomson (the discoverer of the electron). In Thomson's model, an atom consisted of a sphere of positively charged material in which were embedded negatively charged electrons, like chocolate chips in a ball of cookie dough. Consider such an atom consisting of one electron with mass \(m\) and charge \(-e\), which may be regarded as a point charge, and a uniformly charged sphere of charge \(+e\) and radius \(R\). (a) Explain why the electron's equilibrium position is at the center of the nucleus. (b) In Thomson's model, it was assumed that the positive material provided little or no resistance to the electron's motion. If the electron is displaced from equilibrium by a distance less than \(R\), show that the resulting motion of the electron will be simple harmonic, and calculate the frequency of oscillation. (\(Hint:\) Review the definition of SHM in Section 14.2. If it can be shown that the net force on the electron is of this form, then it follows that the motion is simple harmonic. Conversely, if the net force on the electron does not follow this form, the motion is not simple harmonic.) (c) By Thomson's time, it was known that excited atoms emit light waves of only certain frequencies. In his model, the frequency of emitted light is the same as the oscillation frequency of the electron(s) in the atom. What radius would a Thomson-model atom need for it to produce red light of frequency 4.57 \(\times\) 10\(^{14}\) Hz? Compare your answer to the radii of real atoms, which are of the order of 10\(^{-10}\) m (see Appendix F). (d) If the electron were displaced from equilibrium by a distance greater than \(R\), would the electron oscillate? Would its motion be simple harmonic? Explain your reasoning. (\(Historical\) \(note:\) In 1910, the atomic nucleus was discovered, proving the Thomson model to be incorrect. An atom's positive charge is not spread over its volume, as Thomson supposed, but is concentrated in the tiny nucleus of radius 10\(^{-14}\) to 10\(^{-15}\) m.)

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