/*! 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 22 An interstellar dust grain, roug... [FREE SOLUTION] | 91Ó°ÊÓ

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An interstellar dust grain, roughly spherical with a radius of \(3 \times 10^{-7}\) meters, has acquired a negative charge such that its potential is \(-0.15\) volt. How many extra electrons has it picked up? What is the strength of the electric field at its surface, expressed in volts/ meter?

Short Answer

Expert verified
The dust grain has about 1580 extra electrons, and the electric field is approximately 8.92 × 10^6 V/m.

Step by step solution

01

Understanding the Problem

We need to find out the number of excess electrons on a spherical interstellar dust grain that has a known potential, and then calculate the electric field at its surface.
02

Equate Potential to Charge Formula

The potential at the surface of a sphere is given by: \( V = \frac{Q}{4\pi\varepsilon_0 R} \). Here, \( V = -0.15 \) V, \( R = 3 \times 10^{-7} \) m, and \( \varepsilon_0 \) is the vacuum permittivity \( (8.85 \times 10^{-12} \text{ F/m}) \). Rearrange the formula to solve for \( Q \): \( Q = V \cdot 4 \pi \varepsilon_0 R \).
03

Compute the Charge

Substitute the known values into the formula: \( Q = (-0.15) \times 4 \pi \times (8.85 \times 10^{-12}) \times (3 \times 10^{-7}) \). Calculate \( Q \).
04

Calculate Number of Electrons

The charge of one electron is \( e = 1.6 \times 10^{-19} \) C. The number of excess electrons \( n \) is given by \( n = \frac{|Q|}{e} \). Calculate \( n \) using the \( Q \) from Step 3.
05

Calculate Electric Field

The electric field \( E \) at the surface of a sphere is given by \( E = \frac{|Q|}{4\pi\varepsilon_0 R^2} \). Use the charge \( Q \) from Step 3 and calculate \( E \).

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Key Concepts

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

Electrostatics
Electrostatics is a branch of physics that studies electric charges at rest. It involves understanding how charges build up and interact when they are not in motion. In everyday life, static electricity is a common example of electrostatic phenomena. When materials rub against each other, they can transfer electrons and become charged.

In the case of the interstellar dust grain from our exercise, the grain is negatively charged, meaning it has gained extra electrons. The electrostatic potential at the dust grain's surface helps us determine how many additional electrons it holds. By applying the formula for electric potential on the surface of a sphere, we can derive the charge, then calculate the number of excess electrons by dividing the total charge of the grain by the charge of a single electron.

Understanding these calculations can help students grasp how electrostatic principles apply to real-world scenarios, like understanding forces between charged particles or assessing possible impacts of charge buildup.
Electric Field
An electric field is a region around charged particles where other charged particles experience a force. It is a vector field, meaning it has both magnitude and direction, and is commonly measured in volts per meter (V/m).

The electric field's strength at the surface of a charged sphere, like our interstellar dust grain, can be calculated using the formula: \[ E = \frac{|Q|}{4\pi\varepsilon_0 R^2} \] This formula highlights how the field depends on the charge \( Q \) of the grain, the vacuum permittivity \( \varepsilon_0 \), and the radius \( R \) of the sphere.

By calculating the electric field, we understand how strong the force would be on a test charge at the surface of the dust grain. This information is crucial, as electric fields are not only a fundamental concept in electrostatics but are also vital in many practical applications in electronics, engineering, and physics.
Vacuum Permittivity
Vacuum permittivity, denoted as \( \varepsilon_0 \), is a constant that quantifies the ability of a vacuum to permit electric field lines. It is also known as the "electric constant."

In physical terms, \( \varepsilon_0 \) is the measure of how much electric charge is required to generate a unit electric field in a vacuum. This constant plays a key role in Coulomb's law and is crucial in the calculations for both the electric potential and field strength on our charged dust grain.

In our textbook exercise, vacuum permittivity helps in calculating the potential and electric field on a spherical grain by providing a fundamental relationship between the charge and the field it produces. Knowing \( \varepsilon_0 \) allows us to accurately apply formulas to determine electric forces and fields in various physics contexts, providing deeper insights into how charges interact in the absence of any other materials.

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Most popular questions from this chapter

One of two nonconducting spherical shells of radius \(a\) carries a charge \(Q\) uniformly distributed over its surface, the other a charge \(-Q\), also uniformly distributed. The spheres are brought together until they touch. What does the electric field look like, both outside and inside the shells? How much work is needed to move them far apart?

The vector function which follows represents a possible electrostatic field: $$ E_{x}=6 x y \quad E_{y}=3 x^{2}-3 y^{2} \quad E_{z}=0 $$ Calculate the line integral of \(\mathbf{E}\) from the point \((0,0,0)\) to the point \(\left(x_{1}, y_{1}, 0\right)\) along the path which runs straight from \((0,0,0)\) to \(\left(x_{1}, 0,\right.\), 0 ) and thence to \(\left(x_{1}, y_{1}, 0\right)\). Make a similar calculation for the path which runs along the other two sides of the rectangle, via the point \((0,\), \(y_{1}, 0\) ). You ought to get the same answer if the assertion above is true. Now you have the potential function \(\phi(x, y, z)\). Take the gradient of this function and see that you get back the components of the given field.

We have two metal spheres, of radii \(R_{1}\) and \(R_{2}\), quite far apart from one another compared with these radii. Given a total amount of charge \(Q\) which we have to divide between the spheres, how should it be divided so as to make the potential energy of the resulting charge distribution as small as possible? To answer this, first calculate the potential energy of the system for an arbitrary division of the charge, \(q\) on one and \(Q-q\) on the other. Then minimize the energy as a function of \(q\). You may assume that any charge put on one of these spheres distributes itself uniformly over the sphere, the other sphere being far enough away so that its influence can be neglected. When you have found the optimum division of the charge, show that with that division the potential difference between the two spheres is zero. (Hence they could be connected by a wire, and there would still be no redistribution. This is a special example of a very general principle we shall meet in Chapter 3 : on a conductor, charge distributes itself so as to minimize the total potential energy of the system.)

A flat nonconducting sheet lies in the \(x y\) plane. The only charges in the system are on this sheet. In the half-space above the sheet, \(z>0\), the potential is \(\phi=\phi_{0} e^{-k z} \cos k x\), where \(\phi_{0}\) and \(k\) are constants. (a) Verify that \(\phi\) satisfies Laplace's equation in the space above the sheet. (b) What do the electric field lines look like? ( \(c\) ) Describe the charge distribution on the sheet.

To show that it takes more than direction and magnitude to make a vector, let's try to define a vector which we'll name squrl \(\mathbf{F}\) by a relation like Eq. 61 with the right-hand side squared: $$ (\operatorname{squrl} \mathbf{F}) \cdot \hat{\mathbf{i}}=\left[\lim _{a_{i} \rightarrow 0} \frac{\int_{C_{i}} \mathbf{F} \cdot d \mathbf{s}}{a_{i}}\right]^{2} $$ Prove that this does not define a vector. (Hint: Consider reversing the direction of f.).

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