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3. (20 points) Coulomb's Law. In this problem, you will use Coulomb's law, to find the Electric field $\vec{E}(\vec{r})$ above the center of a ring of uniform linear charge density. Assume you have a ring, of radius $r_0$, laying flat in the X-Y plane, centered at the origin, with the z axis perpendicular to it. The ring is of a constant linear charge density $\lambda_0$. The point of interest (where we will calculate the electric field) is a distance $z_0$ along the z-axis, above the origin. (a) (2 points). Draw a diagram of this geometry, marking all axes and named variables above. (b) (2 points). Write Coulomb's Law, in integral vector form. $\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{\vec{q}}{r^2} d\vec{r}$ (c) (6 points). Define $\vec{r}$ (the point of interest), $dq$, and $\vec{r}'$ (the location of $dq$; this will depend on an independent variable $\theta$ over which you will integrate, so give the limits of $\theta$). $\vec{E} = \frac{1}{4\pi\epsilon_0} \int_0^{2\pi} \frac{\lambda_0 r_0 d\theta}{r^2} \vec{r}$ (d) (2 points). Define the vector from the differential charge $dq$ to $\vec{r}$.

          3. (20 points) Coulomb's Law.
In this problem, you will use Coulomb's law, to find the Electric field $\vec{E}(\vec{r})$ above the center of a ring of uniform linear charge density.
Assume you have a ring, of radius $r_0$, laying flat in the X-Y plane, centered at the origin, with the z axis perpendicular to it. The ring is of a constant linear charge density $\lambda_0$. The point of interest (where we will calculate the electric field) is a distance $z_0$ along the z-axis, above the origin.
(a) (2 points). Draw a diagram of this geometry, marking all axes and named variables above.
(b) (2 points). Write Coulomb's Law, in integral vector form.
$\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{\vec{q}}{r^2} d\vec{r}$
(c) (6 points). Define $\vec{r}$ (the point of interest), $dq$, and $\vec{r}'$ (the location of $dq$; this will depend on an independent variable $\theta$ over which you will integrate, so give the limits of $\theta$).
$\vec{E} = \frac{1}{4\pi\epsilon_0} \int_0^{2\pi} \frac{\lambda_0 r_0 d\theta}{r^2} \vec{r}$
(d) (2 points). Define the vector from the differential charge $dq$ to $\vec{r}$.

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3 20 points coulombs law in this problem you will use coulombs law to find the electric field vecevecr above the center of a ring of uniform linear charge density assume you have a ring of r 86831

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University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Frank Deng

06/02/2022

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We want to find the electric field $\vec{E}$ at a point $(0,0,z_0)$ on the z-axis. Show more…

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3. (20 points) Coulomb's Law. In this problem, you will use Coulomb's law, to find the Electric field $\vec{E}(\vec{r})$ above the center of a ring of uniform linear charge density. Assume you have a ring, of radius $r_0$, laying flat in the X-Y plane, centered at the origin, with the z axis perpendicular to it. The ring is of a constant linear charge density $\lambda_0$. The point of interest (where we will calculate the electric field) is a distance $z_0$ along the z-axis, above the origin. (a) (2 points). Draw a diagram of this geometry, marking all axes and named variables above. (b) (2 points). Write Coulomb's Law, in integral vector form. $\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{\vec{q}}{r^2} d\vec{r}$ (c) (6 points). Define $\vec{r}$ (the point of interest), $dq$, and $\vec{r}'$ (the location of $dq$; this will depend on an independent variable $\theta$ over which you will integrate, so give the limits of $\theta$). $\vec{E} = \frac{1}{4\pi\epsilon_0} \int_0^{2\pi} \frac{\lambda_0 r_0 d\theta}{r^2} \vec{r}$ (d) (2 points). Define the vector from the differential charge $dq$ to $\vec{r}$.

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Transcript

00:01 Now here you ask to evaluate the electric field of the foreign charge distribution, right? first you have a ring of charge.

00:05 Let me try to joy it.

00:07 This is a rain of charge, right? and this is the center of the rain, and it's going in this x direction, and like this.

00:14 Okay.

00:16 And this is a wide direction, like this.

00:19 You have a wide direction.

00:21 And this is center, right? and this radius are, and of course you have, and the total charge on this rain, according to the question is q, right, capital q, which is given.

00:36 And you have a small charge, also q here, upon the charge little q here.

00:41 And you ask it, for example, upon p, what the electric field? it's very clear the electric field must be pointing this direction, right? so what we need to do is to, first you're asked to find the electric field due to the rain, right? while the electric field, you can easily perform a simple integration.

01:00 For example, you can look at a small line element here and look at the electric field.

01:06 It generates on this.

01:07 Basically, we just need to look at x direction, x component of the electric field because the wire component is just all can, i mean, the wire component, and actually all of the components perpendicular to the x are canceled each other by symmetry.

01:19 So we just look at the x component.

01:21 So you look at this little element, which i can give it maybe d -seta, right? suppose this angle is d -seta.

01:29 And then you see d -ceta times all right, and that's line element.

01:34 And times the density of the chart, line density of charge that's given by kube over 2 pi hour, right? and then this is the charge.

01:46 And then the electric field, of course, the charge divided by the distance, which is the distance, let's say, let's call this x.

01:54 Okay, then x squared and plus r squared, right? that's the distance squared.

02:00 And of course, you need to look at the x component...

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