Physics, 26.01.2021 02:20 edgarnino13

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of radius R. The central perpendicular axis through the ring is a z-axis, with the origin at the center of the ring. (a) What is the magnitude of the electric field due to the rod at z = 0? N/C

(b) What is the magnitude of the electric field due to the rod at z = infinity? N/C

(c) In terms of R, at what positive value of z is that magnitude maximum? R

(d) If R = 4.00 cm and Q = 9.00 �C, what is the maximum magnitude?N/C

Answers: 1

Show answers

Answers

Answer from: senituliii

a) E_{z} = 0, b) E_{z} = 0, c) z = 1.73 [1 + $\sqrt{1 - \frac{4R^2}{3} }$ ], d) $E_{max}$ Emax = 9.7 10¹⁰ N / C

Explanation:

For this exercise we use the expression

E = k ∫ dq / r²

By applying this expression to our problem of a ring of radius R, with a perpendicular axis in the z direction, we can calculate the electric field for a small charge element

dE = k dq / r²

In the attachment we can see a diagram of the electric field, it is observed that the fields perpendicular to the z axis cancel and the field remains in the direction of the axis

d $E_{z}$ = dE cos φ

we substitute

E_{z} = k∫ $\frac{dq}{r^2}$ cos φ

let's write the expressions

r² = R² + z²

cos φ = z / r

we substitute in the integral, where we see that the load differential does not depend on the distance and the value of the total load is + Q

E_{z} = k $\frac{1}{ (R^2 +z^2) } \ \frac{z}{ (R^2 + z^2)^{1/2} }$ ∫ dq

E_{z} = k Q $\frac{z}{ (R2+z^2)^{3/2} }$

This is the expression for the electric field in the axis perpendicular to the ring, we analyze this expression to answer the questions

a) the magnitude of the field at z = 0

E_{z} = 0

b) the magnitude of the field for z = inf

when z »R the expression remains

E_{z} = k $\frac{z}{z^{3} }$ Q

E_{z} = k Q $\frac{1}{z^2}$

therefore when the value of z = int the field goes to E_{z} = 0

c) In value of z for which the field is maximum.

We have an extreme point when the first derivative is equal to zero

$\frac{dE_z}{dz } = k Q [ (R^2 +z^2)^{3/2} - z \ 3 \frac{z}{ (R^2 +z^2)^{1/2} } = 0$

we solve

(R² + z²)^{ 3/2} = 3 z² /(R² +z²) ^{1/2}

(R² + z²)² = 3z²

r² + z² = √3 z

z² –1.732 z + R² = 0

we solve the quadratic equation

z = [1.732 ± $\sqrt{3 - 4R^2}$ ]/ 2 = [1.73 ± 1.73 $\sqrt{ 1 - \frac{4 R^2}{3} }$ ] / 2

z = 0.865 [1 ± $\sqrt{1 - \frac{4R^2}{3} }$ ]

Therefore there are two points where the field has an extreme point one, one is a maximum and the other a minimum, as we have already determined a minimum at z = 0 the maximum point must be

z = 0.865 [1 + $\sqrt{1 - \frac{4R^2}{3} }$ ]

d) the value of Emax

z₁ = 0.865 [1+ $\sqrt{1 - \frac{4 \ 0.04^2}{3} }$ ) = 1.73 [1 + √0.99786]

z₁ = 1.729 m

z₂ = 0.865 [1 - √0.99786 ]

z₂ = 0.0011 m≈ 0

for which the field has a maximum value substituting in equation 1

$E_{max} = 9 10^9 \ 9 \ \frac{1.729}{(0.04^2 + 1.729^{3/2})}$

$E_{max}$ = 81 10⁹ $\frac{1.729}{1.4408}$

$E_{max}$ Emax = 9.7 10¹⁰ N / C

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of r

Answer from: Quest

Cpowerlifting my boi

Answer from: Quest

Answer (d) the universe had cooled enough for atoms to combine, protons and neutrons were too far apart to form nuclei;

Another question on Physics

Physics, 22.06.2019 13:40

Use the profile illustrated for problem (1). assume that the layer of silty sand fill (sm) is placed over the entire site. estimate the thickness of the fill (h - in feet) that will cause the effective vertical stress at the center of the clay layer to increase by 517.5 pounds per square foot.

Answers: 1

Answer

Physics, 23.06.2019 00:00

You are a researcher for a golf club manufacturer. you are given two identical looking cubes of a metal alloy. you are informed that they are made of the exact same material, but one is crystalline, whereas the other is amorphous. it is your job to determine which one is amorphous this one is more stress-resistant and is useful in reinforcing golf clubs. which of the following is the best way to determine which is which? a. determine the density of each cube. the more dense one is the amorphous solid. b. melt both cubes and look for a broader range of melting temperatures. the one that melts over a broader range of temperatures is the amorphous solid. c. determine the density of each cube. the less dense cube is the amorphous solid. d. melt both cubes and measure the range of melting temperatures. the one that melts over a narrower range of temperatures is the amorphous solid.

Answers: 1

Answer

Physics, 23.06.2019 05:00

Scientists discovered fossils in several layers of the earth you see here. they found fossils of algae, snails, and clams in layer d. given that information, where do you think they found fossil evidence of simple land plants and amphibians?

Answers: 2

Answer

Physics, 23.06.2019 09:30

What is the volume of 2400kg of gasoline (petrol) if the density of petrol is 0.7g/km3

Answers: 1

Answer

You know the right answer?

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of r...

Questions