A charged line like that shown in Fig. 21.25 extends from y=2.50 cm to y=-2.50 cm . The total ch

A charged line like that shown in Fig. 21.25 extends from...

A charged line like that shown in Fig. 21.25 extends from $y=2.50 \mathrm{cm}$ to $y=-2.50 \mathrm{cm} .$ The total charge distributed uniformly along the line is $-9.00 \mathrm{nC}$ . (a) Find the electric field (magnitude and direction) on the $x$ -axis at $x=10.0 \mathrm{cm} .$ (b) Is the magnitude of the electric field you calculated in part (a) larger or smaller than the electric field 10.0 $\mathrm{cm}$ from a point charge that has the same total charge as this finite line of charge? In terms of the approximation used to derive $E=Q / 4 \pi \epsilon_{0} x^{2}$ for a point charge from Eq. $(21.9),$ explain why this is so. (c) At what distance $x$ does the result for the finite line of charge differ by 1.0$\%$ from that for the point charge?

A charged line like that shown in Fig. 21.25 extends from $y=2.50 \mathrm{cm}$ to $y=-2.50 \mathrm{cm} .$ The total charge distributed uniformly along the line is $-9.00 \mathrm{nC}$ . (a) Find the electric field (magnitude and direction) on the $x$ -axis at $x=10.0 \mathrm{cm} .$ (b) Is the
magnitude of the electric field you calculated in part (a) larger or smaller than the electric field 10.0 $\mathrm{cm}$ from a point charge that has the same total charge as this finite line of charge? In terms of the approximation used to derive $E=Q / 4 \pi \epsilon_{0} x^{2}$ for a point charge from Eq. $(21.9),$ explain why this is so. (c) At what distance $x$ does the result for the finite line of charge differ by 1.0$\%$ from that for
the point charge?

Key Concepts

Electric Field from a Continuous Charge Distribution

This concept involves calculating the net electric field produced by a distribution of charge by integrating the contributions from each infinitesimal charge element using Coulomb’s law. It requires establishing a coordinate system, identifying the differential element of charge, and summing the vector contributions over the entire charge distribution to determine the total field.

Uniform Linear Charge Density

Uniform linear charge density is the measure of charge per unit length along a one-dimensional charge distribution. It is calculated by dividing the total charge by the length over which the charge is distributed. This constant density simplifies the integration process when determining the electric field from the line charge.

Coulomb's Law and the Superposition Principle

Coulomb’s law quantifies the electric field due to a point charge and serves as the fundamental basis for calculating the field contributions from small elements of a charge distribution. The superposition principle states that the total electric field generated by a distribution of charges is the vector sum of the fields produced by each charge element, enabling the integration approach used in these problems.

Point Charge Approximation and Far-field Behavior

The point charge approximation treats an extended charge distribution as if all its charge were concentrated at a single point, which is a valid assumption when the observation point is far from the charge distribution compared to its spatial extent. This approximation facilitates comparisons between the fields of distributed and point charges, and provides insight into when and why the electric field deviates from the ideal point charge behavior.

Transcript

00:02 For problem 91, we are given a line charge from negative 2 .5 to 2 .5 cm.

00:08 And the total charge of the line is negative 9 nanoculum.

00:13 So for letter a, we are to find the electric field at x is equal to 10 cm.

00:20 So this is 10 cm.

00:28 And so let's assume this is r, dy, with a distance of y, this d is equal to d squared of point 1 squared plus y squared and this is angle theta and we have a net horizontal force so our e is equal to k q d squared and taking the derivative of this we will have k dq divided by 0 .10 squared plus y squared for our dey, we know that it equals to 0 because the y components would cancel out.

01:26 And for dex we have de, the cosine of theta.

01:34 And the cosine of theta is equals to .1 divided by d square root of .1 squared by squared.

01:45 And our dq is equals to lambda.

01:51 D -y.

01:52 So let's substitute all our values here.

01:56 We have d -e -s, k -lamda, d -y divided by 0 .1 squared plus y squared multiplied by 0 .1 divided by the square of 0 .1 squared, y squared.

02:14 So let's integrate this.

02:17 So we have x is equal to k lambda multiplied by point 1 and then the integral of the y divided by 0 .1 squared plus y squared is 2 3 halves from negative 0 .025 to 0 .0 to 5 and we would get a first let's write what k is and what lambda is so lambda is q 9 times 10 raised to the negative 9 divided by 2a, which is 2 times 0 .025.

03:16 So now we can solve this x...

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