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3 years ago

At what distance along the z-axis is the electric field strength a maximum?

1 answer:

4 0

The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is

dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) 

where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. 

However, cos(T) = x/sqrt(x^2 + R^2) 

so the equation becomes 

dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) 

dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 

Integrating around the ring you get 

E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 

E = (R/2*e0)*x*(x^2 + R^2)^-1.5 

we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is 

dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} 

dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} 

to find the maxima set this = 0, giving 

(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 

mult both side by (x^2 + R^2)^2.5 to get 

(x^2 + R^2) - 3*x^2 = 0 

-2*x^2 + R^2 = 0 

-2*x^2 = -R^2 

x = (+/-)R/sqrt(2)

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A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

2 years ago

The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t

hodyreva [135]

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

$f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}$ .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

$f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}$ .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

7 0

2 years ago

Determine the elastic energy U stored in the compressed spring.

hoa [83]

Answer:

Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.

Explanation:

7 0

2 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

2 years ago

REMARKS The speed found in part (a) is the same as if the woman fell vertically through a distance of 21.9 m. The result of part

sasho [114]

Answer:

Yes, if the system has friction, the final result is affected by the loss of energy.

Explanation:

The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.

In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces

$W_{fr}$ = ΔEm

$W_{fr}$ = $Em_{f}$ -Em₀

As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.

$Em_{f}$ = $W_{fr}$ + Em₀

Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer

½ m v² = -fr d + mgh

v = √(-fr d 2/m + 2 gh)

v = √ (2gh - 2fr d/m)

Now it is clear that there is a decrease in the final body speed.

Consequently, if the system has friction, the final result is affected by the loss of energy.

5 0

2 years ago