/*! 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 36 A point charge of \(-3.0 \times ... [FREE SOLUTION] | 91Ó°ÊÓ

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A point charge of \(-3.0 \times 10^{-5} \mathrm{C}\) is placed at the origin of coordinates in vacuum. Find the electric field at the point \(x=5.0\) \(\mathrm{m}\) on the \(x\) -axis.

Short Answer

Expert verified
The electric field at \(x=5.0\ \mathrm{m}\) is \(1.0788 \times 10^4\ \mathrm{N/C}\) towards the origin.

Step by step solution

01

Understand the problem

We have a point charge of \(-3.0 \times 10^{-5} \mathrm{C}\) located at the origin of a coordinate system. We need to calculate the electric field at a point on the \(x\)-axis, specifically at \(x = 5.0\ \mathrm{m}\).
02

Recall the formula for electric field due to a point charge

The electric field \(E\) due to a point charge \(Q\) at a distance \(r\) is given by the formula: \[ E = \frac{k \cdot |Q|}{r^2} \] where \(k\) is the Coulomb's constant \(8.99 \times 10^9 \ \mathrm{N\cdot m^2/C^2}\).
03

Plug in the given values

We have \(Q = -3.0 \times 10^{-5} \ \mathrm{C}\) and \(r = 5.0\ \mathrm{m}\). Substitute these into the formula: \[ E = \frac{8.99 \times 10^9 \cdot 3.0 \times 10^{-5}}{5.0^2} \].
04

Simplify the expression

Calculate \(5.0^2 = 25.0\), so the electric field becomes: \[ E = \frac{8.99 \times 10^9 \cdot 3.0 \times 10^{-5}}{25} \].
05

Calculate the electric field

Perform the multiplication: \(8.99 \times 10^9 \cdot 3.0 \times 10^{-5} = 2.697 \times 10^5\). Then, divide by 25: \[ E = \frac{2.697 \times 10^5}{25} = 1.0788 \times 10^4 \ \mathrm{N/C} \].
06

Determine the direction of the electric field

Since the charge is negative, the electric field direction is towards the charge. Therefore, at \(x = 5.0\ \mathrm{m}\), the electric field points towards the origin.

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Key Concepts

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

Point Charge
In the world of electricity, the point charge stands out as a fundamental concept. A point charge is a charged particle with a very small size compared to the distances we are examining. This concept simplifies the calculations because the charge can be considered as concentrated at a single point in space. When using the term 'point charge', it often implies that this charge affects its surroundings entirely based on its magnitude and distance from other objects.
The charge is often denoted by the symbol \( Q \), and its magnitude is measured in Coulombs (\( ext{C} \)). In the given exercise, the point charge is \(-3.0 \times 10^{-5} \ ext{C}\) and is located at the origin of the coordinate system. This negative value of the charge will have implications on the direction of the electric field it creates.
Coulomb's Constant
Coulomb's constant, represented by the symbol \( k \), plays a key role in electrostatics, especially in calculating electric fields and forces between charges. It provides the proportionality necessary to relate the electric forces to the charges involved and their separation distance. The accepted value of this constant in a vacuum is \( 8.99 \times 10^9 \ ext{N}\cdot\text{m}^2/\text{C}^2 \).
This constant is derived from Coulomb’s law, which states that the force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. In our exercise, \( k \) is used in the formula to calculate the electric field produced by a point charge. It factors into how strongly the electric field interacts with surrounding points in space.
Electric Field Formula
The electric field formula is incredibly useful for determining the strength of the field generated by a point charge at any position in space. The formula is:
  • \( E = \frac{k \cdot |Q|}{r^2} \)
This equation suggests that the electric field \( E \) depends on several factors:
  • The absolute value of the charge \( Q \), which dictates the magnitude of the field.
  • The inverse square of the distance \( r \) from the charge. This means that as you get farther from the charge, the field gets weaker quite rapidly.
This formula allows us to predict not just the strength but also consider the potential direction of the field when calculating effects due to the charge. Since the charge in the given exercise is negative, although the magnitude is used in calculations, the field direction itself is influenced by the charge's negativity.
Coordinate System
The coordinate system is crucial for determining positions in physical space relating to the point charge. It provides a framework where one can locate the position of the charge and the point where the electric field is being calculated. Generally, this is done in a simple Cartesian coordinate system with an \( x \)-axis, \( y \)-axis, and sometimes \( z \)-axis, but often only one is necessary for simpler problems.
In the current exercise, the point charge is at the origin, which is \((0,0)\) in this system, while the electric field is being calculated at point \((5.0 \ ext{m}, 0)\), meaning the point lies along the positive direction of the \( x \)-axis. Setting up the coordinate system this way simplifies many calculations and visually aids in understanding the problem.
Negative Charge
A negative charge, like the one given in the exercise, influences the direction of the electric field it creates. While positive charges emit electric fields that point outward from the charge, negative charges attract electric field lines inward towards themselves.
In the specific exercise, the charge placed at the origin is \(-3.0 \times 10^{-5} \ ext{C} \). Such a negative sign indicates that any electric field generated will point towards the origin from wherever it's calculated in space. Even though the magnitude of the field was calculated using the absolute value of the charge, the direction is indeed towards the charged particle due to its negative nature. Understanding this can be crucial for predicting the motion of other charged particles in proximity or interpreting the results of more complex electric field scenarios.

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Most popular questions from this chapter

Four equal-magnitude point charges ( \(3.0 \mu\) C) are placed in air at the corners of a square that is \(40 \mathrm{~cm}\) on a side. Two, diagonally opposite each other, are positive, and the other two are negative. Find the force on either negative charge.

Two charged metal plates in vacuum are \(15 \mathrm{~cm}\) apart as drawn in Fig. 24-7. The electric field between the plates is uniform and has a strength of \(E=3000 \mathrm{~N} / \mathrm{C}\). An electron \(\left(q=-e, m_{e}=9.1 \times 10^{-31}\right.\) \(\mathrm{kg}\) ) is released from rest at point \(P\) just outside the negative plate. (a) How long will it take to reach the other plate? (b) How fast will it be going just before it hits? The electric field lines show the force on a positive charge. (A positive charge would be repelled to the right by the positive plate and attracted to the right by the negative plate.) An electron, being negative, will experience a force in the opposite direction, toward the left, of magnitude $$ F_{E}=|q| E=\left(1.6 \times 10^{-19} \mathrm{C}\right)(3000 \mathrm{~N} / \mathrm{C})=4.8 \times 10^{-16} \mathrm{~N} $$ Because of this force, the electron experiences an acceleration toward the left given by $$ a=\frac{F_{E}}{m}=\frac{4.8 \times 10^{-16} \mathrm{~N}}{9.1 \times 10^{-31} \mathrm{~kg}}=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} $$ In the motion problem for the electron released at the negative plate and traveling to the positive plate, $$ v_{i}=0 x=0.15 \mathrm{~m} a=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} $$ (a) Fiom \(x=v_{i} t+\frac{a}{a}\) $$ t=\sqrt{\frac{2 x}{a}}=\sqrt{\frac{(2)(0.15 \mathrm{~m})}{5.3 \times 10^{4^{4}} \mathrm{~m} / \mathrm{s}^{2}}}=2.4 \times 10^{-8} \mathrm{~s} $$ (b) \(v=v_{i}+a t=0+\left(5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\right)\left(2.4 \times 10^{-8} \mathrm{~s}\right)=1.30 \times 10^{7}\) \(\mathrm{m} / \mathrm{s}\) As you will see in Chapter 41, relativistic effects begin to become important at speeds above this. Therefore, this approach must be modified for very fast particles.

A charge of \(+6.0 \mu\) C experiences a force of \(2.0 \mathrm{mN}\) in the \(+x\) direction at a certain point in space. (a) What was the electric field at that point before the charge was placed there? ( \(b\) ) Describe the force a \(-2.0 \mu \mathrm{C}\) charge would experience if it were used instead of the \(+6.0 \mu \mathrm{C}\) charge.

What is the force of repulsion between two argon nuclei that are separated in vacuum by \(1.0 \mathrm{~nm}\left(10^{-9} \mathrm{~m}\right)\) ? The charge on an argon nucleus is \(+18 e\).

Compute \((a)\) the electric field \(E\) in air at a distance of \(30 \mathrm{~cm}\) from a point charge \(q_{\cdot 1}=5.0 \times 10^{-9} \mathrm{C},(b)\) the force on a charge \(q_{\cdot 2}=\) \(4.0 \times 10^{-10} \mathrm{C}\) placed \(30 \mathrm{~cm}\) from \(q_{\cdot 1}\), and \((c)\) the force on a charge \(q_{\cdot 3}=-4.0 \times 10^{-10} \mathrm{C}\) placed \(30 \mathrm{~cm}\) from \(q_{\cdot 1}\) (in the absence of \(\left.q_{\cdot 2}\right)\). (a) \(E=k_{0} \frac{q_{0} \cdot 1}{r^{2}}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{5.0 \times 10^{-9} \mathrm{C}}{(0.30 \mathrm{~m})^{2}}=0.50 \mathrm{kN} / \mathrm{C}\) directed away from \(q_{\cdot 1}\). (b) \(F_{E}=E_{q \cdot 2}=(500 \mathrm{~N} / \mathrm{C})\left(-4.0 \times 10^{-10} \mathrm{C}\right)=2.0 \times \mathrm{N}=0.20 \mu N\) directed away from \(q_{\cdot 1}\). (c) \(F_{E}=E_{q \cdot 3}=(500 \mathrm{~N} / \mathrm{C})\left(-4.0 \times 10^{-10} \mathrm{C}\right)=-0.20 \mu N\) This force is directed toward \(q_{\cdot 1}\).

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