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

Short Answer

Expert verified
The magnitude of the electric field at a point 0.100 m outside the surface of the sphere is approximately \(873.7 N/C\), and at a point 0.100 m below the surface of the sphere is \(0 N/C\).

Step by step solution

01

Understand the problem and identify given values

Given a solid metal sphere with radius \(0.450 m\) and net charge \(0.250 nC\). And, we need to find the electric field at points \(0.100 m\) outside, and \(0.100 m\) inside the sphere. The known values are: \(Q = 0.250 nC = 0.250*10^{-9} C\), \(r1 = 0.450 m + 0.100 m = 0.550 m\) (distance from center for outside point), and \(r2 = 0.450 m - 0.100 m = 0.350 m\) (distance from center of sphere for inside point).
02

Find electric field outside the sphere (E1)

Using Gauss' Law, the electric field outside the sphere (E1) can be found by the formula \(E = {kQ}/{r^{2}}\), where \(k = 8.99*10^{9} N*m^{2}/C^{2}\) (Coulomb's constant). So, the electric field at the outside point is \(E1 = {kQ}/{r_{1}^{2}}= {(8.99*10^{9} N*m^{2}/C^{2}*0.250*10^{-9} C)}/{(0.550 m)^{2}}\).
03

Find electric field inside the sphere (E2)

The electric field inside a conductor in electrostatic equilibrium is always zero, regardless of the position inside the conductor. Therefore, \(E2 = 0 N/C\) at the point 0.100 m below the surface of the sphere.

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Key Concepts

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

Gauss' Law
Gauss' Law is a fundamental concept in electrostatics that enables us to relate electric charges to electric fields. Understanding Gauss' Law can simplify complex problems, especially when dealing with symmetrical objects like spheres. In essence, Gauss' Law states that the total electric flux through a closed surface is directly proportional to the enclosed charge. Mathematically, it is expressed as:\[ \\Phi_E = \oint E \cdot dA = \frac{Q_{enc}}{\varepsilon_0} \\]where \( \Phi_E \) represents the electric flux, \( E \) is the electric field, \( dA \) is a differential area on the Gaussian surface, \( Q_{enc} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space. This law is particularly useful when dealing with spherical symmetry because it allows us to envision a hypothetical "Gaussian Surface" that aligns with the geometry of the problem. In this exercise, the outer point (0.550 m from the center) utilizes Gauss' Law to calculate the electric field emanating from the charged sphere, resulting in a clear understanding of how charges distribute in a field.
Coulomb's constant
Coulomb's constant, often symbolized as \( k \), is a crucial factor in calculations involving electric forces and fields. In the step-by-step solution, you might have noticed it's vital when employing Gauss' Law for electric fields. The value of Coulomb's constant is approximately \( 8.99 \times 10^9 \\mathrm{N\cdot m^2/C^2} \\). It acts akin to the gravitational constant in gravitational force calculations, but for electric charges instead.
Coulomb's constant is derived from the permittivity of free space (\( \varepsilon_0 \)) via the relationship:
- \( k = \frac{1}{4\pi\varepsilon_0} \)- This constant is part of the equation used to find the electric field as one applies Gauss’ Law:
\[ E = \frac{kQ}{r^2} \]
where \( E \) is the magnitude of the electric field at a distance \( r \), and \( Q \) is the point charge. Imperatively, understanding Coulomb's constant allows you to comprehend the strength and character of the electric field created by point charges in electrostatics.
Electrostatic Equilibrium
Electrostatic equilibrium is a state where there is no net motion of charge within a conductor. It is a crucial concept when solving problems involving electric fields within conductors. When a conductor reaches electrostatic equilibrium, several important properties emerge:
  • The electric field inside the conductor is zero.
  • Any excess charge resides on the surface of the conductor.
  • The electric potential is constant throughout the conductor.
In the original exercise, you calculated the electric field within the solid metal sphere and found it to be zero due to electrostatic equilibrium. This demonstrates a fundamental principle: within a conductive material at equilibrium, no electric field can exist inside because any field would cause the free charges to move. Hence, the instantaneous charges redistribute to cancel any internal field, achieving that state of zero electric field. This simplifies problems like the one in the exercise because it allows us to confidently assert that the electric field is zero at any point inside the conductive region of the sphere.

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Most popular questions from this chapter

A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{~m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{~m}\). What is the total electric flux due to the two point charges through a spherical surface centered at the origin and with ra\(\operatorname{dius}(\mathrm{a}) 0.500 \mathrm{~m},(\mathrm{~b}) 1.50 \mathrm{~m},(\mathrm{c}) 2.50 \mathrm{~m} ?\)

A point charge of \(-3.00 \mu \mathrm{C}\) is located in the center of a spherical cavity of radius \(6.50 \mathrm{~cm}\) that, in turn, is at the center of an insulating charged solid sphere. The charge density in the solid is \(\rho=7.35 \times 10^{-4} \mathrm{C} / \mathrm{m}^{3} .\) Calculate the electric field inside the solid at a distance of \(9.50 \mathrm{~cm}\) from the center of the cavity.

A hollow, conducting sphere with an outer radius of \(0.250 \mathrm{~m}\) and an inner radius of \(0.200 \mathrm{~m}\) has a uniform surface charge density of \(+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}\). A charge of \(-0.500 \mu \mathrm{C}\) is now introduced at the center of the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

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

A charge of \(87.6 \mathrm{pC}\) is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of \(29.2 \mathrm{~cm}^{2} .\) A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is \(5.00 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C},\) what area of the sheet is enclosed by the Gaussian surface?

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