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FromTheMoon [43]
3 years ago
8

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

Physics
1 answer:
Lesechka [4]3 years ago
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 

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

<span>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. </span>

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

<span>so the equation becomes </span>

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

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

<span>Integrating around the ring you get </span>

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

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

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

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

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

<span>to find the maxima set this = 0, giving </span>

<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>

<span>mult both side by (x^2 + R^2)^2.5 to get </span>

<span>(x^2 + R^2) - 3*x^2 = 0 </span>

<span>-2*x^2 + R^2 = 0 </span>

<span>-2*x^2 = -R^2 </span>

<span>x = (+/-)R/sqrt(2) </span>
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A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.
Akimi4 [234]

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3.59 m/s

Explanation:

The average velocity is defined as:

\frac{Total displacement}{Total time}

The turtle first walks 18m south, and then is taken 1,1Km (or 1100m) north. Thus, the total displacement  is 1082m north (1100m north - 18m south).

Now we have to calculate the total time, which will be equal to the sum of the time the turtle walked and the time it was taken by truck.

The walking time is 3.5 minutes. Since 1 minute = 60 seconds, then the walking time is 210 seconds.

To calculate the truck time we use the equation:

Time = \frac{Distance}{Speed}

Where the distance the truck travelled is 1100m and the speed of the truck is 12m/s.

Thus,

Truck time= \frac{1100m}{12m/s}= 91.67s

The total time is the sum of the walking time and the truck time.

Total time = 210s + 91.67s = 301.67s.

As mencioned previously, the average velocity is equal to total displacement/ total time, thus:

Average velocity = \frac{1082m}{301.67s} = 3.59 m/s North

Since the average velocity is a vector, it has a magnitude and a direction. In this case the magnitude is 3.59 m/s and the direction is north since the turtle's final displacement is north of where it started.

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