Positive charge Q is distributed uniformly along the x -axis...

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. 21.47$)$ . (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 (magnitnde 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.

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. 21.47$)$ . (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 (magnitnde 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

Coulomb's Law

Coulomb's Law describes the force between two point charges as directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This fundamental law underpins the calculation of electric fields and forces both in discrete and continuous charge configurations.

Electric Field from a Continuous Charge Distribution

Instead of dealing with isolated point charges, a continuous charge distribution requires breaking the distribution into infinitesimal charge elements. The electric field at any point in space is then calculated by integrating the contributions from each infinitesimal element, using Coulomb's Law while taking into account the vector nature of the field.

Superposition Principle

In electromagnetism, the superposition principle states that the total electric field produced by several charge elements is the vector sum of the fields produced by each element independently. This principle is essential when calculating the electric field and resulting forces from a continuous charge distribution.

Integration Techniques in Electrodynamics

Integral calculus is employed to compute the net electric field or force from a continuous distribution of charge. Techniques include setting up the appropriate differential expressions for the field contributions, breaking them into components, and integrating over the spatial region occupied by the charge distribution.

Far-Field (Distant Field) Approximation

When the distance between the observation point and the charge distribution is much greater than the dimensions of the charge distribution, the distributed charge can be approximated as a point charge. This simplification explains why, for large distances, the electric field and force behave similarly to those of a point charge, as indicated by the inverse-square law.

Force on a Charge in an Electric Field

The force on a test charge in an electric field is determined by the product of the test charge and the electric field at its location. This relationship, F = qE, is used to calculate the magnitude and direction of the force exerted by a charge distribution on another charge once the field is determined.

Transcript

00:02 For problem 89, we are given a line charge with a total charge of q and an arbitrary point a plus r, which we have a charge of q.

00:15 So for a, we are to determine the x -y component of the electric field for points on the x axis for x is greater than a.

00:24 So let's set up our x.

00:27 This is rx and a tiny line.

00:31 Which we will call us the x so this would be a plus r minus x and the formula for our electric field is equal kq divided by the distance squared so taking the derivative of this we would have d kdq divided by d squared so we know that lambda is supposed to the q divide by the x so k lambda d x divided by the distance which is a plus r minus x a plus r minus x squared so taking the integral of de this is now d x because our horizontal is the only component of the electric field this would give us k lambda the integral of the x divided the by a plus r minus x squared from 0 to a.

01:48 So evaluating this integral, we would get k lambda multiplied by 1 over a plus r minus x from a to 0.

02:03 So we can get k lambda multiplied by 1 all over r minus 1 over a plus r.

02:16 So substituting we know that our arbitrary point is equals to a plus r.

02:23 Let's call that as our x.

02:26 So our ex is equal to k lambda divided by one all over.

02:36 R is equals to x minus a minus one all over x.

02:43 Let's substitute the k as for pi epsilon not and the lambda as.

02:52 Q divided by a.

02:56 So we would have this equation.

03:03 This is for the electric field horizontal.

03:10 Electric field vertical is equal to zero.

03:15 So for b, we are as the electric force of q on the small q...