A thin rod of length ℓand uniform charge per unit length λlies along the x axis, as shown in Fig

A thin rod of length ℓand uniform charge per unit length...

A thin rod of length $\ell$ and uniform charge per unit length $\lambda$ lies along the $x$ axis, as shown in Figure $\mathrm{P} 23.35 .$ (a) Show that the electric field at $P,$ a distance $y$ from the rod along its perpendicular bisector, has no $x$ component and is given by $E=2 k_{e} \lambda \sin \theta_{0} / y .$ (b) What If? Using your result to part $(\mathrm{a}),$ show that the field of a rod of infinite length is $E=2 k_{e} \lambda / \mathrm{y}$ . (Suggestion: First calculate the field at $P$ due to an element of length $d x,$ which has a charge $\lambda$ dx. Then change variables from $x$ to $\theta,$ using the relationships $x=$ $y \tan \theta$ and $d x=y \sec ^{2} \theta d \theta,$ and integrate over $\theta . )$

A thin rod of length $\ell$ and uniform charge per unit length $\lambda$ lies along the $x$ axis, as shown in Figure $\mathrm{P} 23.35 .$ (a) Show that the electric field at $P,$ a distance $y$ from the rod along
its perpendicular bisector, has no $x$ component and is given by $E=2 k_{e} \lambda \sin \theta_{0} / y .$ (b) What If? Using your result to part $(\mathrm{a}),$ show that the field of a rod of infinite length is
$E=2 k_{e} \lambda / \mathrm{y}$ . (Suggestion: First calculate the field at $P$ due to an element of length $d x,$ which has a charge $\lambda$ dx. Then change variables from $x$ to $\theta,$ using the relationships $x=$ $y \tan \theta$ and $d x=y \sec ^{2} \theta d \theta,$ and integrate over $\theta . )$

Key Concepts

Limit of Infinite Charge Distributions

In many electrostatic problems, taking the limit as the size of the charge distribution approaches infinity provides a simplified formula for the electric field. This approach is essential in understanding the behavior of fields generated by idealized, infinitely extended charge distributions, which can often reveal fundamental characteristics of the system.

Electric Field from Continuous Charge Distributions

This concept involves determining the net electric field by segmenting a continuous charge distribution into infinitesimally small elements, applying Coulomb's law to each element, and then integrating over the entire charge distribution to obtain the cumulative effect.

Application of Coulomb’s Law

Coulomb’s law serves as the foundational principle in electrostatics, providing the force and field due to a point charge. When extended to continuous distributions, it is applied to differential charge elements, enabling the calculation of the overall electric field by integration.

Symmetry in Electric Field Calculations

Symmetry plays a key role in simplifying the computation of electric fields. In configurations where the charge distribution exhibits geometric symmetry, certain directional components of the electric field cancel out, which significantly reduces the complexity of the calculations.

Change of Variables and Trigonometric Substitution

Changing variables within an integral can simplify the process of solving an electric field problem. By substituting the linear coordinate with an angular variable via trigonometric relations, the integration becomes more manageable and often leads to clearer insights into the behavior of the field.

Transcript

00:01 So for this problem we have this rod that is uniformly charged and we have a point p that is this is point p here and we're searching for the electric field, total electric field coming from the rod at the given point p.

00:20 Observe the trigonometric relations.

00:23 So first the electric field at the point b will be the sum of all contributions from all electric fields that are emerging from every dx, so from every small infinitesimal part of the road.

00:40 So we have many small infinitesimal parts of the road like this one right here, everywhere around the road.

00:49 And they are all giving out their electric field, which is very small electric field at the point p.

00:56 Some of all of these small electric fields is the total electric field at the point p, which means, since you're a very small electric field at the point p, which means, summing infinitesimally small length and or function corresponding to them which is the function of electric field corresponding to the small length we're basically integrating we're integrating over the horizontal axis to obtain the final electric field now also notice that this small infinitesimal infinitesimally short part of a thin rod is like a point charge.

01:35 So basically this dx, this red part partition here, can be treated that it is so small that it is like a point charge.

01:46 And the function that gives electric field corresponding to the x is basically k times q over r squared r is this distance here.

01:56 This distance can clearly be related to the x distance and the y distance over the angle.

02:05 And trigonometriculations, which we will use certainly.

02:09 And this q will be, we will have to use dq or definitively small charge corresponding to infinitesimally thin and short part of the rod.

02:21 And this means, i mean, d .e and d .e will be equal to, so constant times.

02:30 And here we have, basically, we will make this a function of the angles.

02:37 So if we can integrate from the theta zero, teta 0 is the angle when we moved corresponding to this length from here to the point b.

02:49 So this wide angle from one side to another side is teta 0 and all the angles in between these two are teta.

02:58 So we will turn this function of distance to be a function of teta.

03:04 And in this case this small length dx will move this will function of dx and of course we know that basically dq over r dq will be equal to lambda dx this is linear density over r which is squared and from the decoran theorem r is basically x minus l half squared plus y squared y well because this is this length from here to here this length is l one half and some minus some length x we get the position of basically this this linked from here to here will be l half minus x so this is why we use that and of course y is a constant we can also use that and this will be also we should notice that all of the all of the fields that are all the components of the electric field that are horizontal that are parallel to the x axis will neglect will cancel each other because the rod is symmetric around the center.

04:53 So only the vertical component will give some contributions.

05:04 So actually at the point p due to the small segment is that it has its horizontal and its vertical components.

05:12 So due to symmetric, the symmetric, the symmetric, the symmetric, the symmetrical, the in this case around this point o, all of the horizontal components will be cancelled.

05:26 So the electric field at the given point in this case, point b will be, will have to be multiplied times, times cosine of x, cosine of theta...

Please give Ace some feedback