Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A charge of \(0.681 \mathrm{nC}\) is placed at \(x=0\). Another charge of \(0.167 \mathrm{nC}\) is placed at \(x_{1}=10.9 \mathrm{~cm}\) on the \(x\) -axis. a) What is the combined electric potential of these two charges at \(x=20.1 \mathrm{~cm},\) also on the \(x\) -axis? b) At which point(s) on the \(x\) -axis does this potential have a minimum?

Short Answer

Expert verified
Answer: To find the combined electric potential at point x = 20.1 cm, use the equation: \(V(20.1 × 10^{-2}) = V_1(20.1 × 10^{-2}) + V_2(20.1 × 10^{-2})\). To find the point(s) where the potential has a minimum value, solve the equation: \(0 = -\frac{k(0.681 × 10^{-9})}{x^2} + \frac{k(0.167 × 10^{-9})}{(x - 10.9 × 10^{-2})^2}\).

Step by step solution

01

Calculate the electric potential due to individual charges

First, let's find the electric potential V(x) for both charges on the x-axis. We'll use the following formula for electric potential: \(V_{i}(x) = \frac{kQ}{r}\), where \(V_{i}(x)\) is the electric potential due to charge \(Q\), \(k\) is the electrostatic constant, \(k = 8.99 × 10^9 N m^2/C^2\), and \(r\) is the distance between the point of interest and the charge. For the two charges, we'll denote their electric potentials as \(V_1(x)\) and \(V_2(x)\). To calculate their electric potentials at any point x: 1) For \(Q_1 = 0.681 nC\): \(r_1 = |x - 0|\) (the distance between the point and the first charge) \(V_1(x) = \frac{kQ_1}{r_1} = \frac{k(0.681 × 10^{-9})}{x}\) 2) For \(Q_2 = 0.167 nC\): \(r_2 = |x - 10.9 × 10^{-2}|\) (the distance between the point and the second charge) \(V_2(x) = \frac{kQ_2}{r_2} = \frac{k(0.167 × 10^{-9})}{x - 10.9 × 10^{-2}}\)
02

Find the combined electric potential

Now, let's find the combined electric potential at the given point x = 20.1 cm. The combined electric potential \(V(x)\) is the sum of the electric potentials due to both charges: \(V(x) = V_1(x) + V_2(x)\). Calculate the two potentials at x = 20.1 cm: \(V_1(20.1 × 10^{-2}) = \frac{k(0.681 × 10^{-9})}{20.1 × 10^{-2}}\) \(V_2(20.1 × 10^{-2}) = \frac{k(0.167 × 10^{-9})}{20.1 × 10^{-2} - 10.9 × 10^{-2}}\) Now, find the combined electric potential: \(V(20.1 × 10^{-2}) = V_1(20.1 × 10^{-2}) + V_2(20.1 × 10^{-2})\)
03

Find the minimum electric potential on the x-axis

To find the point(s) where the potential has a minimum value, we need to find the critical points of the combined potential function \(V(x)\). First, differentiate \(V(x)\) with respect to \(x\): \(\frac{dV}{dx} = \frac{d(V_1(x) + V_2(x))}{dx} = -\frac{k(0.681 × 10^{-9})}{x^2} + \frac{k(0.167 × 10^{-9})}{(x - 10.9 × 10^{-2})^2}\) Now, set \(\frac{dV}{dx} = 0\) to find the critical points: \(0 = -\frac{k(0.681 × 10^{-9})}{x^2} + \frac{k(0.167 × 10^{-9})}{(x - 10.9 × 10^{-2})^2}\) Solve the above equation to obtain the point(s) on the x-axis where the potential has a minimum value.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Electrostatics
Electrostatics is a branch of physics that studies stationary or slow-moving electric charges. It's a fundamental aspect of electricity and is responsible for the forces between charges at rest. An easy way to visualize electrostatics is to think of the classic experiment where a comb, after being run through hair, can pick up small pieces of paper. This demonstrates the presence of static electric charges.

In our exercise, we consider electrostatic interactions between two charges placed at different points along the x-axis. The fascinating part of electrostatics here is how the charges influence the electric potential at different points in space, even without any physical contact. This invisible influence is a characteristic phenomenon of electric fields – regions around a charge where other charges feel a force.
Exploring Coulomb's Law
Coulomb's law is a principle that defines the electrostatic interaction between electrically charged particles. According to this law, the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. In a simpler form, the closer and more charged two objects are, the stronger the force they exert on each other.

In the given exercise, Coulomb's law doesn't directly calculate force, but it's crucial when we apply the concept to find electric potential. The electric potential is related to the work done by or against electrostatic forces, which are governed by Coulomb's law. We can see this relationship in the formula used to calculate electric potential, where the constant 'k' is derived from Coulomb's law, and the charge and distance influence the potential value.
The Role of Electric Charge
Electric charge is a fundamental property of matter expressed in coulombs (C). There are two types of charges: positive and negative. Like charges repel each other, while unlike charges attract. Electric charge is the source of electrostatic forces and electric fields. In our example, we're given two charges, one of 0.681 nC (nanocoulombs) and another of 0.167 nC, which are static charges causing an electric field to emerge around them.

When working with electric charges, it's important to remember they exist even at very small scales. Nanocoulomb, as a unit, helps us deal with the practical quantities of charge we encounter in everyday circumstances.
Deciphering Electric Potential Energy
Electric potential energy is the energy a charged particle holds because of its position in an electric field. Think of it like a rock held up high—it has gravitational potential energy because of its position relative to the Earth. Similarly, a charge in the vicinity of other charges has electric potential energy due to its position relative to the other charges. A fascinating part of electric potential energy is that it allows us to calculate the work that would be involved in moving a charge within the electric field.

The concept of electric potential energy culminates in the idea of electric potential, which is what the exercise asks for. Electric potential, measured in volts, is essentially the electric potential energy per unit charge. Therefore, when talking about the electric potential at a point, we're referring to how much potential energy a unit charge would have at that location. Understanding this concept is key to solving problems like the exercise provided, where we are interested in the combined electric potential due to two separate charges.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A solid conducting sphere of radius \(R\) is charged, and the magnitude of the electric field at the surface of the sphere is \(3.269 \cdot 10^{5} \mathrm{~V} / \mathrm{m} .\) The electric potential \(32.37 \mathrm{~cm}\) from the surface of the sphere is \(2.843 \cdot 10^{5} \mathrm{~V}\). What is the radius, \(R\), of the sphere?

An electron moves away from a proton. Describe how the potential it encounters changes. Describe how its potential energy is changing.

The electron beam emitted by an electron gun is controlled (steered) with two sets of parallel conducting plates: a horizontal set to control the vertical motion of the beam, and a vertical set to control the horizontal motion of the beam. The beam is emitted with an initial velocity of \(2.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s} .\) The width of the plates is \(d=5.00 \mathrm{~cm},\) the separation between the plates is \(D=4.00 \mathrm{~cm},\) and the distance between the edge of the plates and a target screen is \(L=40.0 \mathrm{~cm} .\) In the absence of any applied voltage, the electron beam hits the origin of the \(x y\) -coordinate system on the observation screen. What voltages need to be applied to the two sets of plates for the electron beam to hit a target placed on the observation screen at coordinates \((x, y)=(0 \mathrm{~cm}, 8.00 \mathrm{~cm}) ?\)

One issue encountered during the exploration of Mars has been the accumulation of static charge on land-roving vehicles, resulting in a potential of \(100 .\) V or more. Calculate how much charge must be placed on the surface of a sphere of radius \(1.00 \mathrm{~m}\) for the electric potential just above the surface to be \(100 .\) V. Assume that the charge is uniformly distributed.

A hollow conducting sphere of radius \(R\) is centered about the origin of an \(x y z\) -coordinate system. A total charge \(Q\) is distributed uniformly over the surface of the sphere. Assuming, as usual, that the electric potential is zero at an infinite distance, what is the electric potential at the center of the sphere? a) zero b) \(2 k Q / R\) c) \(k Q / R\) d) \(k Q / 2 R\) e) \(k Q / 4 R\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free