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zloy xaker [14]

3 years ago

at \: is \: amplitude \: {?} " alt="what \: is \: amplitude \: {?} " align="absmiddle" class="latex-formula">

Physics

2 answers:

hichkok12 [17]3 years ago

6 0

<h3><u>Question</u></h3>

what is amplitude?

<h3><u>Answer:</u></h3>

A periodic variable's amplitude is a measure of its change across a single period, such as time or space. Amplitude is defined in a variety of ways, but they're all based on the size of the disparities between the variable's extreme values. The amplitude of a periodic function is sometimes referred to as the phase in older writings texts.

<h3><u>-------------------------</u></h3><h3><u>hope it helps...</u></h3><h3><u>have a great day!!</u></h3>

liubo4ka [24]3 years ago

5 0

Answer:

The amplitude is a measure of its change in a single period (such as time or spatial period)

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2 years ago

A ring of charge of radius r0 lies in the x-y plane with its center at the origin.The charge counterclockwise from θ = 0º to θ =

Sindrei [870]

Answer:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{2QR}{(R^2 + r_0^2)^{3/2}}(-\^y)$

The direction of the electric field is in the negative y-direction.

Explanation:

We will separate the both halves of the ring and calculate the E-field separately. We will use the following formula and the method of integration.

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

We will choose an arbitrary infinitesimal displacement (ds) on the ring. We will apply the above formula to that small segment, then integrate over the half ring.

The small segment has a length ds. This segment is also equal to Rdθ, because any curve on the ring is equal to radius times angle, and the angle of the infinitesimal displacement is the smallest angle we could choose, dθ.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{R^2 + r_0^2}$

dQ can be written in terms of given variables according to the fact that the charge density of the total ring is equal to any small segment of the ring.

$\frac{Q}{\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{\pi}$

dE becomes,

$dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{\pi}\frac{1}{R^2 + r_0^2}$

Before integrating this quantity, we have consider the fact that electric field is a vector, therefore has to be separated into its vertical and horizontal components.

$dE_v = dE\sin{\alpha} = dE \frac{r_0}{\sqrt{R^2 + r_0^2}}\\dE_h = dE\cos{\alpha} = dE \frac{R}{\sqrt{R^2 + r_0^2}}$

These equalities can be deduced from the geometry of the ring.

Now, we can conclude our method by integrating both halves.

$E_1_{v} = \int\limits^{\pi}_0 {dE_1_{v}} \, d\theta = \int\limits^{\pi}_0 {\frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}}\int\limits^{\pi}_0 \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}} \pi$ $E_1_{v} = \frac{1}{4\pi\epsilon_0}\frac{Qr_0}{(R^2 + r_0^2)^{3/2}}\^z$

$E_1_{h} = \int\limits^{\pi}_0 {dE_1_{h}} \, d\theta = \int\limits^{\pi}_0 {\frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}}\int\limits^{\pi}_0 \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}} \pi$ $E_1_{h} = \frac{1}{4\pi\epsilon_0}\frac{QR}{(R^2 + r_0^2)^{3/2}}(-\^y)$

We will repeat these calculations for the bottom half of the ring:

$E_2_{v} = \int\limits^{\pi}_0 {dE_2_{v}} \, d\theta = \int\limits^{\pi}_0 {\frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}}\int\limits^{\pi}_0 \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}} \pi$ $E_2_{v} = \frac{1}{4\pi\epsilon_0}\frac{Qr_0}{(R^2 + r_0^2)^{3/2}}(-\^z)$

$E_2_{h} = \int\limits^{\pi}_0 {dE_2_{h}} \, d\theta = \int\limits^{\pi}_0 {\frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}}\int\limits^{\pi}_0 \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}} \pi$ $E_2_{h} = \frac{1}{4\pi\epsilon_0}\frac{QR}{(R^2 + r_0^2)^{3/2}}(-\^y)$

Therefore, the net electric field created by the ring is the sum of these four components. Note that the vertical components cancel out each other.

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{2QR}{(R^2 + r_0^2)^{3/2}}(-\^y)$

5 0

3 years ago