/*! 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 63 Suppose that to repel electrons ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that to repel electrons in the radiation from a solar flare, each sphere must produce an electric field \(\vec{E}\) of magnitude \(1 \times 10^{6} \mathrm{~N} / \mathrm{C}\) at \(25 \mathrm{~m}\) from the center of the sphere. What net charge on each sphere is needed? (a) \(-0.07 \mathrm{C} ;\) (b) \(-8 \mathrm{mC}\) (c) \(-80 \mu \mathrm{C}\) (d) \(-1 \times 10^{-20} \mathrm{C}\)

Short Answer

Expert verified
The net charge needed on each sphere is closest to option (b) -8 mC

Step by step solution

01

Identify the given parameters

The magnitude of the electric Field \(E\) is given as \(1 x 10^6 N/C\). The distance from the center of the sphere \(r\) is 25m.
02

Write down Coulomb’s Law

Coulomb’s Law \(E= k|Q| / r^2\). In this equation, \(E\) is the electric field, \(k\) is Coulomb's constant, \(Q\) is the charge and \(r\) is the distance from the charge.
03

Rearrange the formula to solve for Q

Rearranging the formula for Q we get \(|Q| = Er^2 / k\)
04

Substitute the given values into the formula and compute Q

Substitute \(E = 1 x 10^6 N/C\), \(r = 25m\) and \(k = 9 x 10^9 Nm^2/C^2\) into the formula. This yields \(|Q| = (1 x 10^6 N/C)x(25m)^2 / (9 x 10^9 Nm^2/C^2) = -2.8 x 10^-3 C = -2.8 mC\)
05

Choose the correct answer

Now compare the calculated result with the provided options. None of the options exactly match but -2.8 mC is closest to option (b) -8 mC

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Key Concepts

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

Coulomb's Law
Understanding Coulomb's Law is pivotal when tackling electric field physics problems. Simply put, Coulomb's Law quantifies the amount of force between two stationary, electrically charged particles. The law's formula is expressed as:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
Here, \( F \) is the electrostatic force (in Newtons), \( k \) is Coulomb's constant (approximately \(8.9875 \times 10^9 N m^2 C^{-2}\)), \( q_1 \) and \( q_2 \) are the magnitudes of electric charges (in Coulombs), and \( r \) is the distance (in meters) separating the charges.

In practice, this formula can be adjusted to find the magnitude of an electric field created by a single charge that's affecting a point in space, leading to the version used in our textbook problem: \[ E = k \frac{|Q|}{r^2} \]
Here, \( E \) represents the electric field magnitude instead of \( F \), and \( Q \) is the charge causing the field. This simplification allows us to isolate the charge we need to determine, to create an electric field of a desired magnitude.
Electric Charge
Moving on to the core concept of 'electric charge', it's critical to recognize that charge is an intrinsic property of matter associated with an electric field. There are two types of electric charges: positive and negative. Like charges repel and opposite charges attract, as seen with commonly known objects such as magnets. The unit for measuring electric charge is the Coulomb, named after Charles-Augustin de Coulomb.

One Coulomb is roughly equivalent to the charge of \(6.242 \times 10^{18}\) protons or electrons. However, it is very rare to encounter such a large amount of charge in practical scenarios. Instead, we often deal with subunits such as milliCoulombs (mC) and microCoulombs (μC), which are suitable for describing typical problem scenarios—like the one in our textbook, which involves the net charge on spheres designed to protect from solar radiations.
Electric Field Magnitude
Lastly, the 'electric field magnitude' is a measure of the electric force per unit charge at a specific point in space. It's represented by \( E \) and measured in Newtons per Coulomb (N/C). The electric field around a charge exists regardless of whether there's another charge feeling its effect or not. Conceptually, it's similar to the gravitational field around a mass.

In our problem, the goal was to find out the electric charge needed on a sphere to produce an electric field with a magnitude of \( 1 \times 10^6 N/C \) at a distance of 25 meters from its center. By using Coulomb’s Law adapted for electric fields, we can relate the desired electric field magnitude to the charge needed to create it. This allows for diverse practical applications, from designing electric shielding to determining the force experienced by particles within various fields.

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Most popular questions from this chapter

In a region of space there is an electric field \(\overrightarrow{\boldsymbol{E}}\) that is in the \(z\) -direction and that has magnitude \(E=[964 \mathrm{~N} /(\mathrm{C} \cdot \mathrm{m})] x\). Find the flux for this field through a square in the \(x y\) -plane at \(z=0\) and with side length \(0.350 \mathrm{~m}\). One side of the square is along the \(+x\) -axis and another side is along the \(+y\) -axis.

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.

You measure an electric field of \(1.25 \times 10^{6} \mathrm{~N} / \mathrm{C}\) at a distance of \(0.150 \mathrm{~m}\) from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius \(0.150 \mathrm{~m} ?\) (b) What is the magnitude of this charge?

A solid insulating sphere has radius \(R\) and carries positive charge distributed throughout its volume. The charge distribution has spherical symmetry but varies with radial distance \(r\) from the center of the sphere. The volume charge density is \(\rho(r)=\rho_{0}(1-r / R)\) where \(\rho_{0}\) is a constant with units of \(\mathrm{C} / \mathrm{m}^{3}\). (a) Derive an expression for the electric field as a function of \(r\) for \(rR .\) (c) At what value of \(r\), in terms of \(R,\) does the electric field have its maximum value?

A square insulating sheet \(80.0 \mathrm{~cm}\) on a side is held horizontally. The sheet has \(4.50 \mathrm{nC}\) of charge spread uniformly over its area. (a) Estimate the electric field at a point \(0.100 \mathrm{~mm}\) above the center of the sheet. (b) Estimate the electric field at a point \(100 \mathrm{~m}\) above the center of the sheet. (c) Would the answers to parts (a) and (b) be different if the sheet were made of a conducting material? Why or why not?

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