/*! 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 37 An electric field has a constant... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electric field has a constant value of \(4.0 \times 10^{3} \mathrm{V} / \mathrm{m}\) and is directed downward. The field is the same everywhere. The potential at a point \(P\) within this region is 155 V. Find the potential at the following points: (a) \(6.0 \times 10^{-3} \mathrm{m}\) directly above \(P\). (b) \(3.0 \times 10^{-3} \mathrm{m}\) directly below \(P,\) (c) \(8.0 \times 10^{-3} \mathrm{m}\) directly to the right of \(P\).

Short Answer

Expert verified
(a) 179 V, (b) 143 V, (c) 155 V.

Step by step solution

01

Understanding the Problem

We have a uniform electric field, directed downward, with magnitude \(E = 4.0 \times 10^3 \text{ V/m}\). The potential at point \(P\) is 155 V. We are asked to find the potential at different points with respect to \(P\).
02

Relationship between Electric Field and Potential

The change in electric potential due to an electric field \(E\) that is linear in nature can be described by \(\Delta V = -E \cdot d\), where \(d\) is the displacement in the direction of the field. If moving downward increases \(d\), the potential decreases.
03

Calculate Potential 6.0 mm Above P

Given that the field is directed downward, moving 6.0 mm (0.006 m) above \(P\) means we move opposite to the field. The change in potential is \(\Delta V = -E \cdot d = -(4.0 \times 10^3) \times (-0.006) = 24 \text{ V}\). Therefore, the potential at this point is \(155 \text{ V} + 24 \text{ V} = 179 \text{ V}\).
04

Calculate Potential 3.0 mm Below P

Moving 3.0 mm (0.003 m) below \(P\), in the direction of the field, we use \(\Delta V = -E \cdot d = -(4.0 \times 10^3) \times 0.003 = -12 \text{ V}\). The potential is \(155 \text{ V} - 12 \text{ V} = 143 \text{ V}\).
05

Calculate Potential 8.0 mm to the Right of P

Since the field direction is not horizontal, moving horizontally by 8.0 mm (0.008 m) to the right of \(P\) does not affect the potential. Therefore, the potential remains the same, \(155 \text{ V}\).

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 91Ó°ÊÓ!

Key Concepts

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

Electric Field
The concept of an electric field is vital in understanding electrical forces and potentials. Think of an electric field as an invisible map of forces around charged objects, pushing or pulling on other charged entities. The electric field strength, measured in volts per meter (V/m), signifies how strong this force is in a particular direction. For example, if you have a field strength of \(4.0 \times 10^{3} \, \mathrm{V/m}\), it means that there is a force exerting a push or pull equivalent to 4,000 volts for every meter in that direction. In this setup, the field is directed downward. Therefore, any positive charge would feel a downward force, moving along the field lines.

An electric field can affect how electric potential, or voltage, changes with position. The direction of the electric field is key in determining whether electric potential increases or decreases as you move between points. Remember, the electric field lines show the path charges would take if free to move, from areas of high potential energy to low, similar to a ball rolling downhill.
Potential Difference
Understanding potential difference is crucial for comprehending how electrical energy is transferred or transformed. The potential difference, or voltage difference, is the energy needed to move a unit charge between two points in an electric field. It's calculated using the formula \(\Delta V = -E \cdot d\), where \(\Delta V\) is the change in potential, \(E\) is the electric field strength, and \(d\) is the distance moved along the field lines.

In our exercise, moving 6.0 mm above point \(P\) is against the direction of the field, hence involving a positive change in potential, whereas moving 3.0 mm below \(P\) along the field results in a decrease in potential. When moving perpendicular to the field direction, such as 8.0 mm to the right, the potential remains unchanged, emphasizing that potential difference heavily depends on movement relative to field lines.
Uniform Electric Field
Learning about uniform electric fields helps us predict how electric potential changes across different spaces. A uniform electric field is effectively constant in magnitude and direction at every point. This characteristic simplifies calculations, allowing us to predict potential differences using linear algebra. In such fields, the relationship \(\Delta V = -E \cdot d\) holds no matter where you are, provided you stay within the field.

In practical terms, a uniform field means identical conditions across space, making it easier to calculate potential changes. For example, in our exercise, the field is directed downward, remaining the same throughout the region. Whether you move 6.0 mm up, 3.0 mm down, or even laterally, calculations remain straightforward owing to this uniformity. This predictability ensures efficient problem-solving and aids practical applications, such as in design fields involving capacitors or simulating electric fields in technology.

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 parallel plate capacitor has a capacitance of \(7.0 \mu \mathrm{F}\) whenfilled with a dielectric. The area of each plate is \(1.5 \mathrm{m}^{2}\) and the separation between the plates is \(1.0 \times 10^{-5} \mathrm{m} .\) What is the dielectric constant of the dielectric?

The capacitance of an empty capacitor is \(1.2 \mu \mathrm{F}\). The capacitor is connected to a \(12-\mathrm{V}\) battery and charged up. With the capacitor connected to the battery, a slab of dielectric material is inserted between the plates. As a result, \(2.6 \times 10^{-5} \mathrm{C}\) of additional charge flows from one plate, through the battery, and onto the other plate. What is the dielectric constant of the material?

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of \(\operatorname{mass} 7.20 \times 10^{-3} \mathrm{kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

The electric potential energy stored in the capacitor of a defibrillator is 73 J, and the capacitance is \(120 \mu \mathrm{F}\). What is the potential difference that exists across the capacitor plates?

Two identical point charges \(\left(q=+7.20 \times 10^{-6} \mathrm{C}\right)\) are fixed at diagonally opposite corners of a square with sides of length \(0.480 \mathrm{m}\). A test charge \(\left(q_{0}=-2.40 \times 10^{-8} \mathrm{C}\right)\), with a mass of \(6.60 \times 10^{-8} \mathrm{kg},\) is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square.

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.