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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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What is the velocity of an object that has a momentum of 4,000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.
julia-pushkina [17]
<h2>Answer: 34.78 m/s</h2>

Explanation:

The momentum p is given by the following equation:

p=m.V   (1)

Where:

m is the mass of the object

V is the velocity of the object

Finding the velocity from (1):

V=\frac{p}{m}   (2)

V=\frac{4000kg.m/s}{115kg}  

<u>Finally:</u>

V=34.78m/s >>>This is the velocity of the object

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Three 1.5V cells are connected in series in a circuit. What is the total potential difference?
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Answer:

4.5V

Explanation:

1.5x3= 4.5

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True or false The cardiovascular system includes the heart and blood vessels.
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The answer to this question is false
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Gravity is the correct answer.

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In a lab, four balls have the same velocities but different masses.
olya-2409 [2.1K]

Answer:

New Momentum of Ball B=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}

<u>Explanation:</u>

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2\frac{m}{s}

Momentum of Ball A=2.2\frac{k g m}{s}

Momentum of Ball B=4.4 \frac{k g m}{s}

Momentum of Ball C=11\frac{k g m}{s}

Momentum of Ball D=15\frac{k g m}{s}

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2\frac{m}{s}

Initial Momentum of Ball B=4.4\frac{k g m}{s}

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

                                            =3×2=6kg

Thus we get Mass of Ball B=6kg

According to the formula,  

Change in momentum of Ball B \Delta p=m \times \Delta v

Where \Delta p=change in momentum

          m=mass of the ball B

         \Delta v=change in velocity ball B

And \Delta v=v, since all balls, have same velocity

Thus the above equation, changes to

         \Delta p=m \times v

Substitute all the values in the above equation we get

         \Delta p=6 \times 2.2

                     =13.2 \frac{\mathrm{kgm}}{\mathrm{s}}  

Result:

 Thus the New Momentum of ball B=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}

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3 years ago
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