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Chapter 26: Problem 33

Surrounding an electrically charged particle is an electric field. The rate of change of electric potential with respect to the distance from the particle creating the field equals the negative of the value of the electric field. That is, \(d V / d x=-E,\) where \(E\) is the electric field. If \(E=k / x^{2},\) where \(k\) is a constant, find the electric potential at a distance \(x_{1}\) from the particle, if \(V \rightarrow 0\) as \(x \rightarrow \infty\)

Short Answer

Expert verified
The electric potential is given by \( V(x_1) = -\frac{k}{x_1} \).

Step by step solution

01

Understanding the Problem

We need to find the electric potential function \( V(x) \) given the relationship \( \frac{dV}{dx} = -E \) and \( E = \frac{k}{x^2} \). The condition \( V \to 0 \text{ as } x \to \infty \) will help determine the integration constant.
02

Substituting the Electric Field Expression

Substitute \( E = \frac{k}{x^2} \) into the differential equation: \( \frac{dV}{dx} = -\frac{k}{x^2} \). This equation can now be integrated with respect to \( x \).
03

Integrating with Respect to Distance

Integrate \( \frac{dV}{dx} = -\frac{k}{x^2} \) with respect to \( x \). This gives \( V(x) = \int -\frac{k}{x^2} \, dx \).
04

Solving the Integral

Calculate the integral: \[V(x) = \int -\frac{k}{x^2} \, dx = k \left(-\frac{1}{x}\right) + C = -\frac{k}{x} + C.\]Here, \( C \) is a constant to be determined.
05

Applying the Boundary Condition

Use the given boundary condition \( V \to 0 \text{ as } x \to \infty \). This means that the constant \( C \) must be zero for the potential to approach zero at infinity. Thus, \( V(x) = -\frac{k}{x} \).
06

Expressing the Final Solution

Now that we have determined \( C = 0 \), the electric potential at a distance \( x_1 \) from the particle is given by:\[V(x_1) = -\frac{k}{x_1}.\]

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Key Concepts

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

Electric Field
An electric field is a region around a charged particle where other charged particles experience a force. The field can be thought of as a "map" showing the direction and magnitude of this force at any given point. The strength of the electric field, often denoted by \(E\), is typically measured in volts per meter (V/m). It provides crucial information about how a charged particle would move if placed within the field.
  • The electric field can be visualized as arrows pointing away from positively charged particles and toward negatively charged ones.
  • In equations, \(E\) represents the electric field intensity at a specific point.
  • It is defined such that the force \(F\) on a charge \(q\) is given by \(F = qE\).
Understanding electric fields is important for designing electronic devices and understanding phenomena in physics.
Rate of Change
The rate of change in calculus refers to how a quantity changes with respect to another. In the context of electric fields and potentials, it describes how the electric potential changes as you move through the field.
  • The rate of change of electric potential with respect to distance \(x\) is expressed as the derivative \(\frac{dV}{dx}\).
  • For electric fields, this rate of change gives the negative of the electric field: \(\frac{dV}{dx} = -E\).
This relationship highlights that as the distance from a charged particle increases, the electric potential typically decreases, reflecting a weakening field.
Distance
Distance in the context of electric fields usually refers to the separation between a point of interest and the source of the electric field. It plays a critical role in determining the strength of the field and the potential at a point.
  • In the formula \(E = \frac{k}{x^2}\), distance \(x\) directly affects the value of the electric field \(E\) at a point.
  • The farther away you are from a charge, the weaker the electric field becomes, reflecting the inverse square law dependency.
Understanding distance's impact is crucial for calculating electric potential or fields, especially when plotting potential on graphs or designing circuits.
Integration in Calculus
Integration is a fundamental concept in calculus that allows us to find the original function from its derivative. In physical contexts like electric fields, it is used to determine quantities like electric potential from known changes (or derivatives) such as the electric field.
  • In this exercise, we integrate \(\frac{dV}{dx} = -\frac{k}{x^2}\) to find \(V(x)\).
  • The integral of \(-\frac{k}{x^2}\) is calculated as \(-\frac{k}{x} + C\), where \(C\) is the constant of integration.
Integration helps us translate dynamic changes into static values, essential for engineering and scientific calculations involving forces and potentials.

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Most popular questions from this chapter

Find the work done in pumping the water out of the top of a cylindrical tank 3.00 ft in radius and 10.0 ft high, given that the tank is initially full and water weighs \(62.4 \mathrm{lb} / \mathrm{ft}^{3}\). (Hint: If horizontal slices \(d x\) ft thick are used, each element weighs \(62.4(\pi)\left(3.00^{2}\right) d x\) Ib, and each element must be raised \(10-x \mathrm{ft}\), if \(x\) is the distance from the base to the element (see Fig. 26.70 ). In this way, the force, which is the weight of the slice, and the distance through which the force acts are determined. Thus, the products of force and distance are summed by integration.)

Find the indicated moment of inertia or radius of gyration. Find the radius of gyration of a plate covering the region bounded by \(x=2, x=4, y=0,\) and \(y=4,\) with respect to the \(y\) -axis.

Some applications of areas are shown. Because the displacement \(s,\) velocity \(v,\) and time \(t\) of a moving object are related by \(s=\int v d t,\) it is possible to represent the change in displacement as an area. A rocket is launched such that its vertical velocity \(v\) (in \(\mathrm{km} / \mathrm{s}\) ) as a function of time \(t\) (in s) is \(v=1-0.01 \sqrt{2 t+1} .\) Find the change in vertical displacement from \(t=10 \mathrm{s}\) to \(t=100 \mathrm{s}.\)

Find the force on one side of a cubical container \(6.0 \mathrm{cm}\) on an edge if the container is filled with mercury. The density of mercury is \(133 \mathrm{kN} / \mathrm{m}^{3}\)

Find the areas bounded by the indicated curves, using (a) vertical elements and (b) horizontal elements. $$y=8 x, x=0, y=4$$
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