Find the electric field a distance z above one end of a straight line segment of length L (Fig.

Find the electric field a distance z above one end of a...

Key Concepts

Coulomb's Law

Coulomb's Law is the fundamental principle that describes the electric force between two point charges. It forms the basis for calculating electric fields from charge distributions by relating the force to the product of the charges and the inverse square of the distance between them. This law is indispensable when extending the concept to continuous distributions of charge, where the overall field is obtained by integrating the contributions of infinitesimal charge elements.

Continuous Charge Distribution

A continuous charge distribution refers to the spread of electric charge over a region of space, rather than being confined to discrete points. In problems involving a line, surface, or volume charge distribution, the uniformity or variation of the charge density directly influences how the electric field is calculated. Here, a uniform line charge simplifies the integration process since the charge per unit length is constant along the segment.

Integration in Electrostatics

Integration is a key technique used to sum the infinitesimal contributions of electric fields generated by each element of a continuous charge distribution. This process involves setting up an integral that accounts for both the magnitude and the direction of the field contributions from each element, often requiring careful consideration of the geometry and limits of integration.

Limiting Behavior and Asymptotic Analysis

Verifying the limiting behavior of a derived formula is crucial for ensuring its consistency with physical expectations. In electrostatics, analyzing the case where the observation distance is much larger than the dimensions of the charge distribution (z >> L) helps confirm that the field approaches that of a simpler system, such as a point charge. This asymptotic analysis acts as a sanity check for the more general, derived expression.

Transcript

00:01 Hi there, so for this problem we are asked about to find the electric field at distance seat above one end of a straight line sediment of length held as is shown in this figure, which carries an uniform light charge lambda.

00:18 We need to check that your formula is consistent with what you will expect for the case of seed much greater than else.

00:32 Now, the situation that we have in this case is shown in here.

00:38 What we have drowned in here is the electric field that is produced by an infinitely segment of this line of charge that we can express as the differential of charge is equal to the the charge density, the uniform line charge, density times the differential in adds.

01:08 Now with that said we know that this is going to produce an electric field company in the c direction that is going to be 1 over 4 times pi times epsilon sub 0 times the integral from 0 to 0 because that is the total distance of landa times the differential in x times the radius in here, this separation to the square, this separation in here, and this cosine of theta, because that is the seed component.

01:48 Now in here, if you use trigonometry, you will find that r square is equal to cd square plus x square, and also by using trigonometry, we know that cosine of f squared, theta is seed divided by the radius r.

02:12 So what we are going to do is to substitute this definition in here.

02:20 So we are going to have that now.

02:22 This component of the electric field is 1 over 4 times pi times epsilon sub 0 times lambda times seed because we, can take that out and then what we are this seed comes from the fat from the cosine from this cosine function of theta now these times the integral from zero to l of one over c2 square plus x square and also this elevated to 3 over 2 times the differential in x.

03:31 Now we know that the integral of this form is simple, and that is going to be the following.

03:45 Times lambda, times seed, times 1 over seed square, and this times x divided by the square root of adds c to the square plus x to the square, and this evaluated from zero to l.

04:10 Then the solution for this is going to be equal to lambda, divided by 4 times pi times epsilon sub -0 times c, and this times l divided by the square root of cid square plus l square.

04:29 So this is the seed component of the electric field.

04:36 Now for the x component of the electric field, we do something similar...

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