A total charge of Q is distributed uniformly on a line...

A total charge of $Q$ is distributed uniformly on a line segment of length $2 L$ along the $y$ -axis (see figure). The $x$ -component of the electric field at a point $(a, 0)$ is given by $$E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}}$$ where $k$ is a physical constant and $a>0$ a. Confirm that $E_{x}(a)=\frac{k Q}{a \sqrt{a^{2}+L^{2}}}$ b. Letting $\rho=Q / 2 L$ be the charge density on the line segment, show that if $L \rightarrow \infty,$ then $E_{x}(a)=2 k \rho / a$ (See the Guided Project Electric field integrals for a derivation of this and other similar integrals.)

A total charge of $Q$ is distributed uniformly on a line segment of length $2 L$ along the $y$ -axis (see figure). The $x$ -component of the electric field at a point $(a, 0)$ is given by
$$E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}}$$
where $k$ is a physical constant and $a>0$
a. Confirm that $E_{x}(a)=\frac{k Q}{a \sqrt{a^{2}+L^{2}}}$
b. Letting $\rho=Q / 2 L$ be the charge density on the line segment, show that if $L \rightarrow \infty,$ then $E_{x}(a)=2 k \rho / a$ (See the Guided Project Electric field integrals for a derivation of this and other similar integrals.)

Key Concepts

Limiting Behavior in Charge Distribution Problems

Analyzing the limiting behavior of a physical system, such as considering an infinitely long charge distribution, provides insight into the asymptotic form of the electric field. This approach helps simplify the integration process and reveals the field’s dependence on parameters such as distance.

Definite Integral Evaluation in Electromagnetism

Evaluating definite integrals is central in electromagnetism problems, particularly when calculating fields due to continuous charge distributions. Techniques such as substitution and recognizing standard integral forms are crucial for analytically solving these integrals.

Continuous Charge Distribution

When charge is spread uniformly along a length, area, or volume, it is described by a charge density (linear, surface, or volume). Understanding how to express a total charge in terms of a density and a differential element is key to setting up and evaluating integrals for electric field calculations.

Coulomb's Law

Coulomb's Law explains the force and field between charges, and in the context of continuous charge distributions, it serves as the foundation for calculating the electric field at a point due to infinitesimal charge elements. Each charge element contributes to the total field, and integrating these contributions yields the field for an extended object.

Electric Field Integration and Superposition

The principle of superposition states that the net electric field is the vector sum of the fields produced by individual charge elements. This concept requires integrating over the entire charge distribution, taking into account both the magnitude and direction of the field contributions.

Transcript

00:01 In this question, it is given that a charge q is distributed uniformly over a line segment of length to l, which is along y axis, and electric field component on the x axis is given by the expression, which is x of a equals to k multiplied by q multiplied by a over twice of l.

00:23 Integral of dy over a square plus y square to the power 3 by 2 from minus l to l is the limit.

00:31 So first of all, we in the first part, we need to confirm that the electric field at point a will be equal to kq over a times square root a square plus l square.

00:51 So first of all, we will find out the indefinity integral which is dy over a square, plus y square to the power 3 by 2.

01:00 So the integral can be find out by this is the integral dy over a square plus y square to the power 3 by 2.

01:21 So first of all we will substitute y as a tangent theta and therefore dy will be equals to a dot secant square theta dot d theta and substituting this we will get integral i as integration of a secant square theta dot d theta over a square plus y square will be replaced by a square tangent square theta to the power 3 by 2 and by simplifying this we will get this integral as i equal to integral of a second square theta dot d theta over a square can be taken out of this backer and this will become a square to the power 3 by 2 which is a cube.

02:27 1 plus tangent square theta to the power 3 by 2.

02:33 And we know that 1 plus tangent square theta is second square theta.

02:36 So this integral will become integration of there will be second square theta dot d theta over.

02:47 Here a by a cube will be only a square here and this plus 1 plus a tangent square theta is second square theta to the power three by two will give second cube theta so by simplifying this we will get the integral i equals to integration second theta second square theta over second cube theta will be one over second theta which is cosine theta so this will be uh one over a square can be written out of this integral so one over second theta is cosine theta dot d theta we know the integral of cosine theta is sine theta so this will become 1 over a square sine theta now we will replace this sine theta by the expression in terms of y as we substituted y equals to a tangent theta from here we will get tangent theta equals to y over a now we will draw a right angle triangle in which perpendicular is y and base is a so we drawn that right angle triangle here the perpendicular is y and base is a and by pythagros theorem we can find out this hypotenus as a square plus y square whole under the root now we can replace sin theta by perpendicular over hypodice which is y over square root a square plus y square and this integral will be equal to 1 over a square times y over square root a square plus y square now we got the integral now we will find out the value of electric field at point a as the as per the given expression the given expression is e x of a is equal to k times k times a over twice of l an integral for the limits minus l to l and here is dy over square root a square plus y square, sorry, there is no square root.

05:28 This is a square plus y square to the power 3 by 2...

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