CALC Positive charge Q is distributed uniformly along the x...

CALC Positive charge $Q$ is distributed uniformly along the $x$ -axis from $x=0$ to $x=a$ . A positive point charge $q$ is located on the positive $x$ -axis at $x=a+r,$ a distance $r$ to the right of the end of $Q$ (Fig. P21.89). (a) Calculate the $x$ - and y-components of the electric field produced by the charge distribution $Q$ at points on the positive $x$ -axis where $x>a$ . (b) Calculate the force (magnitude and direction) that the charge distribution $Q$ exerts on $q$ . (c) Show that if $r \gg a$ , the magnitude of the force in part (b) is approximately $Q q / 4 \pi \epsilon_{0} r^{2}$ . Explain why this result is obtained.

CALC Positive charge $Q$ is distributed uniformly along the $x$ -axis from $x=0$ to $x=a$ .
A positive point charge $q$ is located on the positive $x$ -axis at $x=a+r,$ a distance $r$ to the
right of the end of $Q$ (Fig. P21.89). (a) Calculate the $x$ - and y-components of the electric field
produced by the charge distribution $Q$ at points on the positive $x$ -axis where $x>a$ . (b) Calculate the force (magnitude and direction) that the charge distribution $Q$ exerts on $q$ . (c) Show that if $r \gg a$ , the magnitude of the force in part (b) is approximately $Q q / 4 \pi \epsilon_{0} r^{2}$ .
Explain why this result is obtained.

Key Concepts

Electric Field

The electric field is a vector field that represents the force experienced per unit charge at any point in space due to a source charge or collection of charges. It is a fundamental concept in electromagnetism and is calculated using Coulomb's law for point charges or by integrating over a continuous charge distribution, ensuring that the directional properties and spatial variations are properly accounted for.

Continuous Charge Distribution

A continuous charge distribution describes how electric charge is spread out over a region of space, which can be along a line, over a surface, or throughout a volume. Instead of working with discrete charges, the distribution is broken into infinitesimal elements where the charge is assumed to be uniformly present, and the overall effect is computed by integration over the entire distribution, capturing the cumulative influence of all the charge elements.

Coulomb's Law

Coulomb's Law is the fundamental principle that quantifies the electric force between two charged objects. It states that the magnitude of the force between point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance separating them. This law is adapted in the context of continuous charge distributions by integrating the contribution of each infinitesimal charge element to determine the overall force or field.

Integration in Electrostatics

Integration is an essential mathematical tool used in electrostatics to sum the contributions from infinitesimally small charge elements when dealing with continuous charge distributions. By integrating the differential contributions (such as small electric fields or forces) across the entire charge distribution, one can derive the net electric field or force at a given point, accounting for variations in distance and direction of each element.

Far-field Approximation

The far-field approximation is a method used when the observation point is at a distance much greater than the dimensions of the charge distribution. Under this condition, the distribution appears nearly point-like, and the resulting electric field or force approximates that due to a single point charge. This simplification is useful in reducing complex integrals to more manageable forms and provides insight into the behavior of the field at large distances.

Vector Decomposition

Vector decomposition is the process of breaking a vector into its components along specified coordinate axes, typically to simplify calculations in problems involving directionality. In electrostatics, decomposing the electric field or force vectors into their x-, y-, and z-components allows for a clearer analysis and integration of contributions from different parts of a distributed charge, making it easier to apply symmetry arguments and solve the resulting equations.

Transcript

00:02 In this question, we have this situation.

00:05 So along the x -axis, there's a rod, positively charge, length a, k, 0 to a, and then there's also a positive charge, plus q, this is plus big q, and then this is at a plus r, okay, and then there are three parts in this question, basically we want to find a, electric field due to the charge rod, the electric force on charge q, the point charge q, and then what happens when the distance between the two charge objects gets further and further away.

00:51 Okay? all right, so let's get started.

00:56 In prior, you want to find the electric field due to the charge rod at any point x along the x -axis.

01:05 Okay.

01:07 So, um, so we have a continuous charge distribution and then based on the situation we can immediately say that there's no y component in the x direction so let's say that's there's a point x here so the electric field is going to be pointing to the right okay yeah the eq2 due to the last q so the y component is going to be zero because all the small little charges in the rod is going to exert and produce an electric fuel that's pointing to the right at point x.

01:47 So suppose we have this point x and then here will be the distance will be x minus a and then this distance will just be x.

01:59 Okay yeah we want to note this distance because we are going to break the rod into small little pieces and each piece has charged dq and then is equal to lambda d x and lambda is the charge density linear charge density okay and lambda is linear charge density since a charge is uniformly distributed okay so we just have q over a okay and then um so e x is going to be the integral of uh 1 over 4 pi epsilon not uh d q over x square okay okay so this and then our dq is lambda the x and we are going to integrate from x minus a to x okay and then our dq is lambda the x over x square then we put out the lambda k if the lambda is q over a okay and then we have one over x square the x squared the x integration q over 4 pi epsilon or a minus 1 over x x x minus a and then put in the limits we are going to have um 1 over x minus 1 over x minus 1 over x okay so this is the answer for part a okay yeah so this is the x and then we have the e y a if the x and y the x and the x and of the electric fuel due to the charge rod at point x...

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