Potentials Along Two Paths

Classical Mechanics Level 3

A force field is described as follows:

F ⃗ = r ^ x + y \vec{F} =\frac{\hat{r}}{ x + y}

The attached diagram shows two paths from P ⃗ 1 = ( 1 , 0 ) \vec{P}_1 = (1,0) to P ⃗ 2 = ( 2 , 3 ) \vec{P}_2 = (2,3) . Define two potential differences:

Δ U 1 = ∫ C 1 F ⃗ ⋅ d ℓ ⃗ Δ U 2 = ∫ C 2 F ⃗ ⋅ d ℓ ⃗ \Delta U_1 = \int_{C_1} \vec{F} \cdot \vec{d \ell} \\ \Delta U_2 = \int_{C_2} \vec{F} \cdot \vec{d \ell}

What is Δ U 1 Δ U 2 \large{\frac{\Delta U_1}{\Delta U_2}} ?

Details and Assumptions
1) r ⃗ = ( x , y ) \vec{r} = (x,y) . r ^ \hat{r} is a unit-length version of r ⃗ \vec{r}
2) Note that the coordinates of P ⃗ 1 \vec{P}_1 have changed relative to the previous problem


The answer is 0.8439.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Nov 19, 2019

Line integral along C1:

C 1 : y = 3 ( x − 1 ) C1: y = 3(x-1)

d l ⃗ = ( i ^ + 3 j ) d x d\vec{l} = (\hat{i} + 3{j})dx

F ⃗ = x i ^ + y j ^ ( x + y ) x 2 + y 2 = x i ^ + 3 ( x − 1 ) j ^ ( 4 x − 3 ) x 2 + 9 ( x − 1 ) 2 \vec{F} = \frac{x \hat{i} + y\hat{j}}{(x+y)\sqrt{x^2 + y^2}}=\frac{x \hat{i} + 3(x-1)\hat{j}}{(4x-3)\sqrt{x^2 + 9(x-1)^2}}

I 1 = ∫ C 1 F ⃗ ⋅ d l ⃗ = ∫ 1 2 ( 10 x − 9 ) d x ( 4 x − 3 ) x 2 + 9 ( x − 1 ) 2 ≈ 0.9487 I_1 = \int_{C1} \vec{F} \cdot d\vec{l} = \int_{1}^{2} \frac{(10x-9)dx}{(4x-3)\sqrt{x^2 + 9(x-1)^2}} \approx 0.9487

Now, to evaluate the line integral along C2, which comprises of two branches:

Along the line y = 0 y = 0 :

d l ⃗ = d x i ^ d\vec{l} = dx \hat{i}

F ⃗ = x i ^ + y j ^ ( x + y ) x 2 + y 2 = i ^ x \vec{F} = \frac{x \hat{i} + y\hat{j}}{(x+y)\sqrt{x^2 + y^2}}=\frac{\hat{i}}{x}

I 21 = ∫ y = 0 F ⃗ ⋅ d l ⃗ = ∫ 1 2 d x x = ln ⁡ 2 I_{21} = \int_{y=0} \vec{F} \cdot d\vec{l} = \int_{1}^{2} \frac{dx}{x} = \ln{2}

Along the line x = 2 x = 2 :

d l ⃗ = d y j ^ d\vec{l} = dy \hat{j}

F ⃗ = x i ^ + y j ^ ( x + y ) x 2 + y 2 = 2 i ^ + y j ^ ( 2 + y ) 4 + y 2 \vec{F} = \frac{x \hat{i} + y\hat{j}}{(x+y)\sqrt{x^2 + y^2}}= \frac{2 \hat{i} + y\hat{j}}{(2+y)\sqrt{4+y^2}}

I 22 = ∫ x = 2 F ⃗ ⋅ d l ⃗ = ∫ 0 3 y d y ( 2 + y ) 4 + y 2 ≈ 0.431 I_{22} = \int_{x=2} \vec{F} \cdot d\vec{l} = \int_{0}^{3} \frac{ydy}{(2+y)\sqrt{4+y^2}} \approx 0.431

The required answer is:

I 1 I 21 + I 22 ≈ 0.8439 \boxed{\frac{I_1}{I_{21}+I_{22}} \approx 0.8439}

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