Voltage Difference Along a Path

[Jake 7174]

Homework Statement

Given the electric field E = (2x−y2)ax + (3z−2xy)ay + 3yaz, and the piecewise linear path joining the points A(−2,1,−1), P(2,1,−1), Q(2,3,−1) and B(2,3,1), find −∫E⋅dl from A to P − ∫E⋅dl from P to Q − ∫E⋅dl from Q to B along the straight line segments.

Homework Equations

−∫E⋅dl

(x - x₁) / (x₁ - x₂ ) = (y - y₁) / (y₁ - y₂ ) =(z - z₁) / (z₁ - z₂ )

The Attempt at a Solution

I have solved this problem going direct from point a to b by using the above equation and found that the answer is 8 v. This is confirmed by the answer key. Here is my problem, when I set up the line equation to go from point A to P I get the following;

(x + 2) / (-2 - 2) = (y - 1) / (1 - 1) = (z + 1) / (-1 + 1)

or, (x + 2) / -4 = (y - 1) / 0 = (z + 1) / 0

I am ultimatley going to solve for y and z in terms of x to express my integral in terms of just x and I know how to get rid of the pesky dy and dz as well but I need the line equation to not blow up on me.

How do I resolve the division by 0? I know that the V_AQ = 4 v from the solution but I am not sure what to do about the line equation.

I have an exam on Monday and I am certain a question like this will appear.

 

[TSny]

Here is my problem, when I set up the line equation to go from point A to Q

Did you mean to write A to P?

I get the following;

(x + 2) / (-2 - 2) = (y - 1) / (1 - 1) = (z + 1) / (-1 + 1)

or, (x + 2) / -4 = (y - 1) / 0 = (z + 1) / 0

I am ultimatley going to solve for y and z in terms of x to express my integral in terms of just x and I know how to get rid of the pesky dy and dz as well but I need the line equation to not blow up on me.

How do I resolve the division by 0?

The division by 0 comes from the special nature of the path from A to P. For that path, how would you simplify the expression

##\vec{E}\cdot d\vec{l} = E_xdl_x+E_ydl_y+E_zdl_z##?

 

[Jake 7174]

Did you mean to write A to P?The division by 0 comes from the special nature of the path from A to P. For that path, how would you simplify the expression

##\vec{E}\cdot d\vec{l} = E_xdl_x+E_ydl_y+E_zdl_z##?

Did you mean to write A to P?The division by 0 comes from the special nature of the path from A to P. For that path, how would you simplify the expression

##\vec{E}\cdot d\vec{l} = E_xdl_x+E_ydl_y+E_zdl_z##?

I did mean to write A to P. I fixed it.

if y = z then dl_y = dl_z = 0

so the expression simplifies to ##\vec{E}\cdot d\vec{l} = E_xdl_x##

 

[Jake 7174]

I did mean to write A to P. I fixed it.

if y = z then dl_y = dl_z = 0

so the expression simplifies to ##\vec{E}\cdot d\vec{l} = E_xdl_x##

If this is the case then I have
-∫ (2x - y^2) dx from x= -2 to 2 but Y =1

so, -∫ (2x - 1^2) dx = 4

This is the result I am looking for. I want to be sure I didnt luck into it and that I simplified it correctly as you suggested.

 

[TSny]

If this is the case then I have
-∫ (2x - y^2) dx from x= -2 to 2 but Y =1

so, -∫ (2x - 1^2) dx = 4

This is the result I am looking for. I want to be sure I didnt luck into it and that I simplified it correctly as you suggested.

That looks very good.

 

[Jake 7174]

That looks very good.

Awesome, Thanks for your help