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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.

Short Answer

Expert verified
(a) 7.39 N/C outside; (b) 0 N/C inside.

Step by step solution

01

Understanding the Problem

We have a metal sphere with a radius of 0.450 m carrying a net charge of 0.250 nC. We need to find the electric field at two different points. Firstly, at 0.100 m outside the sphere's surface and secondly, at 0.100 m inside the sphere's surface.
02

Using Gauss's Law for Outside Field Calculation

For a point outside a charged sphere, the sphere can be considered a point charge located at its center. The electric field at any distance, r, from the center is determined using Gauss's law:\[ E = \frac{kQ}{r^2} \]where \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\), \(Q\) is the net charge, and \(r\) is the distance from the center of the sphere to the point, which is the sphere's radius plus the distance outside the surface.
03

Calculate Total Distance from the Center for Outside Point

The radius of the sphere is 0.450 m, and the point is 0.100 m outside the surface, so the total distance from the center is:\[ r = 0.450 + 0.100 = 0.550 \, \text{m} \].
04

Calculate Electric Field at 0.550 m Outside

Plug in the values into the formula for electric field:\[ E = \frac{(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)(0.250 \times 10^{-9} \, \text{C})}{(0.550 \, \text{m})^2}\]Calculating this gives:\[ E \approx 7.39 \, \text{N/C} \].
05

Understanding the Electric Field Inside a Conductor

For the point inside the conducting sphere, the electric field is zero because the electric charges redistribute themselves across the surface of the sphere, canceling the field inside in a static situation.
06

Conclusion

For a point outside the sphere (0.550 m from the center), the electric field magnitude is \(7.39\, \text{N/C}\). For a point inside the sphere, the electric field is \(0 \, \text{N/C}\) by the property of conductors at electrostatic equilibrium.

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Key Concepts

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

Gauss's Law
Gauss's Law is a powerful principle used in electromagnetism to relate the electric field distribution to the charge distribution within a closed surface. According to Gauss's Law, the total electric flux through a closed surface is proportional to the enclosed charge. Mathematically, this can be expressed as:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a vector representing an infinitesimal area on the closed surface, \( Q_{enc} \) is the total charge enclosed within the surface, and \( \varepsilon_0 \) is the permittivity of free space.
This concept is particularly useful for symmetrical charge distributions, like a charged sphere, as it simplifies calculations significantly by allowing us to treat the sphere as a point charge, especially when calculating the field outside the sphere.
  • Only the charge enclosed by the Gaussian surface affects the electric field.
  • The symmetry of the charge distribution simplifies the application of Gauss's Law.
metal sphere charge distribution
In the context of a metal sphere, charge distribution is an interesting topic because of the manner charges spread across conductive materials. When a metal sphere is charged, the charges (typically electrons) move freely until equilibrium is reached. In electrostatic equilibrium:
  • All excess charges reside on the surface of the conductor.
  • The electric field inside a conductor is zero when in a steady state.
This occurs because the electric charges repulse each other and try to get as far away from each other as possible, leading to a surface concentration. For a conducting sphere, such as in this exercise, the inside electric field becomes zero. This happens as the charges rearrange in such a way that they cancel any internal field.
The charges being on the surface is why the calculation for electric field inside the sphere typically yields zero, as stated in the step-by-step solution. Since there are no charges inside to contribute to a field, the net field within remains zero.
Coulomb's constant
Coulomb's constant, denoted as \( k \), is a value that arises from Coulomb's Law, which describes the electrostatic interaction between electric charges. It is defined as:
\[ k = 8.99 \times 10^{9} \, \text{Nm}^2/\text{C}^2 \]Coulomb's Law uses this constant to describe the force between two charges, and it also applies when calculating electric fields around charged objects:
\[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the force between two point charges \( q_1 \) and \( q_2 \), and \( r \) is the distance between them.
This constant is crucial in calculating the electric field using Gauss's Law outside the sphere. By treating the charged sphere as a point charge at its center, we effectively apply Coulomb's constant to find the electric field at any point outside the sphere, as per the problem solution. This is possible because of the symmetry of the sphere.

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Most popular questions from this chapter

How many excess electrons must be added to an isolated spherical conductor 26.0 cm in diameter to produce an electric field of magnitude 1150 N/C just outside the surface?

A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho\). (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho\). (b) What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r = R\). (d) Graph the electric-field magnitude as a function of r from \(r = 0\) to \(r = 3R\).

Which statement is true about \(\overrightarrow{E}\) inside a negatively charged sphere as described here? (a) It points from the center of the sphere to the surface and is largest at the center. (b) It points from the surface to the center of the sphere and is largest at the surface. (c) It is zero. (d) It is constant but not zero.

A 6.20-\(\mu\)C point charge is at the center of a cube with sides of length 0.500 m. (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides were 0.250 m long? Explain.

A region in space contains a total positive charge \(Q\) that is distributed spherically such that the volume charge density \(\rho(r)\) is given by $$\rho(r) = 3ar/2R \space \space \space \space \space \mathrm{for} \space r \leq R/2$$ $$\rho(r) = \alpha[1-(r/R)^2] \space \space \space \space \mathrm{for} \space R/2 \leq r \leq R$$ $$\rho(r) = 0 \space \space \space \space \space \space \mathrm{for} \space r \geq R$$ Here \(\alpha\) is a positive constant having units of C/m\(^3\). (a) Determine \(\alpha\) in terms of \(Q\) and \(R\). (b) Using Gauss's law, derive an expression for the magnitude of the electric field as a function of \(r\). Do this separately for all three regions. Express your answers in terms of \(Q\). (c) What fraction of the total charge is contained within the region \(R/2 \leq r \leq R\)? (d) What is the magnitude of \(\overrightarrow{E}\) at \(r = R/2\)? (e) If an electron with charge \(q' = -e\) is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why?

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