Line Integral + Potential Difference Problem Need

In summary: You will be left with an integral in terms of theta \Delta V = - \frac{.4}{\sqrt{2}} \int_1 ^2 d \theta = - \frac{.4}{\sqrt{2}} \theta \vert_1 ^2 You will have to do a little work to find the appropriate limits for your other line integral.
  • #1
desibabu90
11
0

Homework Statement


The cyclotron is a device for accelerating charged particles. It requires changing electric fields and a magnetic field, but it can be modeled using (non-physical) static electric fields and potentials as we will do in this problem. Keep in mind that these fields cannot actually be created by any set of static charges, so just treat the field as given. To aid in visualization of the problem, consider that this problem could also have been titled, "The Spiral Staircase."

Consider an electric field defined by
E = 0.4*(-sin([tex]\theta[/tex]) [tex]\hat{i}[/tex]+ cos([tex]\theta[/tex]) [tex]\hat{j}[/tex])/r
(E in N/C, i is the unit vector along the x-axis, j is the unit vector along the y-axis; the coefficient 0.40 varies from person to person)

Note that we can write any position in the xy-plane in a similar way:
r = r(cos([tex]\theta[/tex]) i + sin([tex]\theta[/tex]) j ).
The Dot product of E and r would equal 0.

We want to find the potential difference between the two points: (2.7,1.3) and (4.0, 4.9).

a. First integrate radially outward along a line of constant theta. What's the electric potential difference between the points (2.7,1.3) and (5.70,2.74)?

b. Now integrate along a circle of constant radius. What's the potential difference between the points (5.70,2.74) and (4.0,4.9)?

c. What's the total potential difference along the above path between the points (2.7,1.3) and (4.0,4.9)?

Now reverse the order of integration.

d. First integrate along a circle of constant radius. What's the potential difference between the points (2.7,1.3) and (1.90,2.32)?

e. Now integrate along a line of constant theta. What's the potential difference between the points (1.90,2.32) and (4.0,4.9)?

f. What's the total potential difference along this path between the points (2.7,1.3) and (4.0,4.9)?

Now consider two points at the same radius, but on opposite sides of the circle: (2.7,1.3) and (-2.7,-1.3)

g. What's the potential difference between (2.7,1.3) and (-2.7,-1.3) integrating along a circle of constant radius, going in the positive theta direction.

h. What's the potential difference integrating along a circle in the negative theta direction?

Homework Equations


I know you have to do a line integral

[tex]\Delta[/tex]V = -[tex]\int[/tex][tex]\vec{E}[/tex][tex]\vec{ds}[/tex]


The Attempt at a Solution



I got part a and e they both are zero because they are going to be on equipotential lines and therefore there won't a potential difference

however i can't seem to do a line integral.

for part b, i have to parametrize the curve with the points given,
r(t) = (1 -t)<2.7,1.3> + t*(1.9,2.32)
r(t) = <2.7 - .8*t, 1.3 - 1.02*t)
r'(t) = (-.8,-1.02)
but now I don't know how to add it into the integral.

Please help.

thanks in advanced
 
Last edited:
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  • #2
desibabu90 said:

Homework Statement


The cyclotron is a device for accelerating charged particles. It requires changing electric fields and a magnetic field, but it can be modeled using (non-physical) static electric fields and potentials as we will do in this problem. Keep in mind that these fields cannot actually be created by any set of static charges, so just treat the field as given. To aid in visualization of the problem, consider that this problem could also have been titled, "The Spiral Staircase."

Consider an electric field defined by
E = 0.4*(-sin([tex]\theta[/tex]) [tex]\hat{i}[/tex]+ cos([tex]\theta[/tex]) [tex]\hat{j}[/tex])/r
(E in N/C, i is the unit vector along the x-axis, j is the unit vector along the y-axis; the coefficient 0.40 varies from person to person)

Note that we can write any position in the xy-plane in a similar way:
r = r(cos([tex]\theta[/tex]) i + sin([tex]\theta[/tex]) j ).
The Dot product of E and r would equal 0.

We want to find the potential difference between the two points: (2.7,1.3) and (4.0, 4.9).

a. First integrate radially outward along a line of constant theta. What's the electric potential difference between the points (2.7,1.3) and (5.70,2.74)?

b. Now integrate along a circle of constant radius. What's the potential difference between the points (5.70,2.74) and (4.0,4.9)?

c. What's the total potential difference along the above path between the points (2.7,1.3) and (4.0,4.9)?

Now reverse the order of integration.

d. First integrate along a circle of constant radius. What's the potential difference between the points (2.7,1.3) and (1.90,2.32)?

e. Now integrate along a line of constant theta. What's the potential difference between the points (1.90,2.32) and (4.0,4.9)?

f. What's the total potential difference along this path between the points (2.7,1.3) and (4.0,4.9)?

Now consider two points at the same radius, but on opposite sides of the circle: (2.7,1.3) and (-2.7,-1.3)

g. What's the potential difference between (2.7,1.3) and (-2.7,-1.3) integrating along a circle of constant radius, going in the positive theta direction.

h. What's the potential difference integrating along a circle in the negative theta direction?

Homework Equations


I know you have to do a line integral

[tex]\Delta[/tex]V = -[tex]\int[/tex][tex]\vec{E}[/tex][tex]\vec{ds}[/tex]


The Attempt at a Solution



I got part a and e they both are zero because they are going to be on equipotential lines and therefore there won't a potential difference

however i can't seem to do a line integral.

for part b, i have to parametrize the curve with the points given,
r(t) = (1 -t)<2.7,1.3> + t*(1.9,2.32)
r(t) = <2.7 - .8*t, 1.3 - 1.02*t)
r'(t) = (-.8,-1.02)
but now I don't know how to add it into the integral.

Please help.

thanks in advanced

Here's how I would do your line integral [tex] \Delta V = -\int \vec{E} \cdot d \vec{S} [/tex]

First, remember that your path is along a circular arc from (5.70, 2.74) to (4.0, 4.9). Express
[tex] d \vec{S} [/tex] in terms of distance along that arc as

[tex] d \vec{S} = \hat{\theta} r d \theta [/tex]

The unit vector [tex] \hat{\theta} [/tex] is given by

[tex] \hat{\theta} = \frac{-sin(\theta)}{\sqrt{2}} \hat{i} + \frac{cos( \theta)}{\sqrt{2}} \hat{j} [/tex]

You will then need the angle (in radians) between the x-axis and the line from the origin to the point (5.70, 2.74) and the angle (in radians) between the x-axis and the line from the origin to the point (4.0, 4.9). Those two angles will be the limits of your integration. Your integral now becomes

[tex] \Delta V = - \int_1 ^2 \frac{.4}{r}( -sin(\theta) \hat{i} + cos(\theta) \hat{j} ) \cdot (\frac{ -sin(\theta)}{\sqrt{2}} \hat{i} + \frac{cos(\theta)}{\sqrt{2}} \hat{j}) r d \theta [/tex]

Notice the good things that happen: The r's cancel and the trig functions will disappear when you take the dot product. That will make the integral rather easy.
 
  • #3
AEM said:
Here's how I would do your line integral [tex] \Delta V = -\int \vec{E} \cdot d \vec{S} [/tex]

First, remember that your path is along a circular arc from (5.70, 2.74) to (4.0, 4.9). Express
[tex] d \vec{S} [/tex] in terms of distance along that arc as

[tex] d \vec{S} = \hat{\theta} r d \theta [/tex]

The unit vector [tex] \hat{\theta} [/tex] is given by

[tex] \hat{\theta} = \frac{-sin(\theta)}{\sqrt{2}} \hat{i} + \frac{cos( \theta)}{\sqrt{2}} \hat{j} [/tex]

You will then need the angle (in radians) between the x-axis and the line from the origin to the point (5.70, 2.74) and the angle (in radians) between the x-axis and the line from the origin to the point (4.0, 4.9). Those two angles will be the limits of your integration. Your integral now becomes

[tex] \Delta V = - \int_1 ^2 \frac{.4}{r}( -sin(\theta) \hat{i} + cos(\theta) \hat{j} ) \cdot (\frac{ -sin(\theta)}{\sqrt{2}} \hat{i} + \frac{cos(\theta)}{\sqrt{2}} \hat{j}) r d \theta [/tex]

Notice the good things that happen: The r's cancel and the trig functions will disappear when you take the dot product. That will make the integral rather easy.


ok.. so got r's canceling out, and trigs just turn into [tex] .4*(1/{\sqrt{2}}) [/tex] once you dot them. so will the integral be just [tex] - \int .4*(1/{\sqrt{2}}) d\theta [/tex] & for the bounds for part b would it be from 1 to 2 or arctan(2.74/5.7) to arctan(4.9/4)

thanx for ur help
 
  • #4
while doing what i described above i got my answer as -.2828
tat was not the right answer. Did I do something wrong? Please help!
 
  • #5
desibabu90 said:
while doing what i described above i got my answer as -.2828
tat was not the right answer. Did I do something wrong? Please help!

Well, without seeing the details of your work, I can't do anything more than guess where you might have made a mistake. I just calculated out the answer to be -1.24 . One question: is your calculator in radians mode, or degrees mode? That's a common mistake.
 

Related to Line Integral + Potential Difference Problem Need

1. What is a line integral?

A line integral is a mathematical concept used in vector calculus to calculate the total value of a vector field along a given curve. It takes into account both the direction and magnitude of the vector at each point on the curve.

2. How is a line integral related to potential difference?

A line integral can be used to calculate the potential difference between two points in an electric field. This is done by integrating the electric field along the path between the two points, taking into account the direction of the field and the distance traveled.

3. What is the significance of solving line integral + potential difference problems?

Solving these types of problems is important in understanding the behavior of electric fields and their impact on charged particles. It also allows for the calculation of important quantities such as work done and potential energy.

4. What are some common applications of line integral + potential difference problems?

These types of problems are commonly used in engineering and physics, particularly in the analysis of electric circuits and electromagnetic fields. They can also be applied in practical situations, such as calculating the voltage drop across a resistor in a circuit.

5. What are some useful techniques for solving line integral + potential difference problems?

Some useful techniques for solving these types of problems include using the fundamental theorem of calculus, identifying conservative vector fields, and using parametrization to simplify the integration process. It is also important to carefully consider the direction and orientation of the curve and the vector field being integrated.

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