1. Vector Notation
  2. Tangent Vectors
  3. Normal Vectors
  4. Notation and Vector Fields
  5. Field Lines and Equipotentials

Due on Thursday, September 2nd at 3:00 pm.

Send your solutions to the email address rweigel+phys305@gmu.edu as a scanned PDF. If you do not have a scanner, use a smartphone PDF scanning app such as Genius Scan or Adobe Scan. Do not send photos of your assignment. If you have difficulty with this, feel free to contact me for help.

You are welcome to ask questions about homework problems on Discord. I will not give direct answers, but I will give hints. I will also first ask you for your solution to a related problem that was covered in class or in the notes. You are welcome to answer questions from other students on Discord, but please only give suggestions and hints.

This homework involves topics covered in Vectors, Vector Fields, Field Lines, and Equipotentials.

1 Vector Notation

Given r \mathbf{r} is a vector from the origin to the point ( x , y , z ) (x,y,z) and r β€² \mathbf{r}' is a vector from the origin to the point ( x β€² , y β€² , z β€² ) (x',y',z') and a scalar function f f defined according to

f = 1 ∣ r βˆ’ r β€² ∣ 2 f = \frac{1}{|\mathbf{r}-\mathbf{r}'|^2}

  1. Write f f in cartesian coordinates.

  2. Show that ∣ r βˆ’ r β€² ∣ 2 |\mathbf{r}-\mathbf{r}'|^2 can be written as r 2 + r β€² 2 βˆ’ 2 r β‹… r β€² r^2+r'^2-2\mathbf{r}\cdot\mathbf{r}'

  3. Compute βˆ‡ ( 1 ∣ r βˆ’ r β€² ∣ ) \boldsymbol{\nabla}\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) , where βˆ‡ f ( x , y , z ) = βˆ‚ f βˆ‚ x x ^ + βˆ‚ f βˆ‚ y y ^ + βˆ‚ f βˆ‚ z z ^ \boldsymbol{\nabla}f(x,y,z)=\frac{\partial f}{\partial x}\mathbf{\hat{x}}+\frac{\partial f}{\partial y}\mathbf{\hat{y}}+\frac{\partial f}{\partial z}\mathbf{\hat{z}}

  4. If Τ‰ ≑ r βˆ’ r β€² \textbf{\char"0509} \equiv \mathbf{r}-\mathbf{r}'

    a. Show that

    Τ‰ ^ 2 Τ‰ 2 = ∣ r βˆ’ r β€² ∣ 3 ∣ r βˆ’ r β€² ∣ 3 \frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}=\frac{\phantom{|}\mathbf{r}-\mathbf{r}'\phantom{|^3}}{|\mathbf{r}-\mathbf{r}'|^3}

    b. Write Τ‰ ^ / Τ‰ 2 \hat{\textbf{\char"0509}}/{\char"0509^2} in cartesian coordinates with cartesian unit vectors.

Answer

A common issue was that students dropped the vector symbols. For example, instead of writing 2 r βƒ— β‹… r βƒ— β€² 2\vec{r}\boldsymbol{\cdot}\vec{r}' , they wrote 2 r β‹… r β€² 2r\boldsymbol{\cdot} r' . In the second equation, the dot could be interpreted as a multiplication symbol. If this is the case, then both equations yield a scalar, but the answer will not always be the same. If the dot is interpreted as indicating a dot product, then 2 r β‹… r β€² 2r\boldsymbol{\cdot} r' does not make sense because the dot product always involves two vectors.

  1. f = 1 ( x βˆ’ x β€² ) 2 + ( y βˆ’ y β€² ) 2 + ( z βˆ’ z β€² ) 2 \displaystyle f = \frac{1}{(x-x')^2+(y-y')^2+(z-z')^2}

  2. There are several ways of doing this.

    1. Use the law of cosines. If Ο† \varphi is the angle between two lines of length r r and r β€² r' with both having an endpoint at the origin, then from the law of cosines, C 2 = r 2 + r β€² 2 βˆ’ 2 r r β€² cos ⁑ Ο† C^2=r^2 + r'^2 - 2rr'\cos\varphi .

      From the definition of the dot product, 2 r r β€² cos ⁑ Ο† = 2 r β‹… r β€² 2rr'\cos\varphi=2\mathbf{r}\boldsymbol{\cdot} \mathbf{r}' , giving

      C 2 = r 2 + r β€² 2 βˆ’ 2 r β‹… r β€² C^2=r^2+r'^2-2\mathbf{r}\boldsymbol{\cdot} \mathbf{r}' .

      In this equation C 2 = ∣ r βˆ’ r β€² ∣ 2 C^2 = |\mathbf{r}-\mathbf{r}'|^2 from vector addition, so finally

      ∣ r βˆ’ r β€² ∣ 2 = r 2 + r β€² 2 βˆ’ 2 r β‹… r β€² |\mathbf{r}-\mathbf{r}'|^2=r^2+r'^2-2\mathbf{r}\boldsymbol{\cdot} \mathbf{r}'

    2. Use ∣ r βˆ’ r β€² ∣ 2 = ( r βˆ’ r β€² ) β‹… ( r βˆ’ r β€² ) |\mathbf{r}-\mathbf{r}'|^2=(\mathbf{r}-\mathbf{r}')\boldsymbol{\cdot} (\mathbf{r}-\mathbf{r}') with r = x x ^ + y y ^ + z z ^ \mathbf{r}=x\mathbf{\hat{x}} + y\mathbf{\hat{y}} + z\mathbf{\hat{z}} and r β€² = x β€² x ^ + y β€² y ^ + z β€² z ^ \mathbf{r}'=x'\mathbf{\hat{x}} + y'\mathbf{\hat{y}} + z'\mathbf{\hat{z}} . This gives

      x 2 + y 2 + z 2 + x β€² 2 + y β€² 2 + z β€² 2 βˆ’ 2 ( x x β€² + y y β€² + z z β€² ) = r 2 + r β€² 2 βˆ’ 2 r β‹… r β€² x^2+y^2+z^2 + x'^2+y'^2+z'^2 - 2(xx' + yy' + zz') = r^2+r'^2-2\mathbf{r}\boldsymbol{\cdot}\mathbf{r}'

    3. Use ∣ r βˆ’ r β€² ∣ 2 = ( r βˆ’ r β€² ) β‹… ( r βˆ’ r β€² ) |\mathbf{r}-\mathbf{r}'|^2=(\mathbf{r}-\mathbf{r}')\boldsymbol{\cdot} (\mathbf{r}-\mathbf{r}') . This expands to r β‹… r βˆ’ 2 r β‹… r β€² + r β€² β‹… r β€² = r 2 + r β€² 2 βˆ’ 2 r β‹… r β€² \mathbf{r}\boldsymbol{\cdot}\mathbf{r}-2\mathbf{r}\boldsymbol{\cdot}\mathbf{r}'+\mathbf{r}'\boldsymbol{\cdot}\mathbf{r}'=r^2+r'^2-2\mathbf{r}\boldsymbol{\cdot}\mathbf{r}' . You can get the correct answer using sloppy notation and this method. Some students wrote that the dot product gives r 2 + r β€² 2 βˆ’ 2 r r β€² r^2+r'^2-2rr' and then replaced 2 r r β€² 2rr' with 2 r β‹… r β€² 2\mathbf{r}\boldsymbol{\cdot}\mathbf{r}' , which is not valid.

  1.  

    Let g = 1 ∣ r βˆ’ r β€² ∣ = [ ( x βˆ’ x β€² ) 2 + ( y βˆ’ y β€² ) 2 + ( z βˆ’ z β€² ) 2 ] βˆ’ 1 / 2 \displaystyle g = \frac{1}{|\mathbf{r}-\mathbf{r}'|} =\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{-1/2}

    x ^ βˆ‚ g βˆ‚ x = x ^ βˆ‚ βˆ‚ x [ ( x βˆ’ x β€² ) 2 + ( y βˆ’ y β€² ) 2 + ( z βˆ’ z β€² ) 2 ] βˆ’ 1 / 2 \displaystyle \mathbf{\hat{x}}\frac{\partial g}{\partial x} =\mathbf{\hat{x}}\frac{\partial}{\partial x} \left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{-1/2}

    Using the chain rule, this is

    x ^ βˆ‚ g βˆ‚ x = βˆ’ 1 2 2 ( x βˆ’ x β€² ) [ ( x βˆ’ x β€² ) 2 + ( y βˆ’ y β€² ) 2 + ( z βˆ’ z β€² ) 2 ] 3 / 2 x ^ \displaystyle \phantom{ \mathbf{\hat{x}}\frac{\partial g}{\partial x}} = \frac{-\frac{1}{2}2(x-x')}{\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{3/2}}\mathbf{\hat{x}}

    x ^ βˆ‚ g βˆ‚ x = βˆ’ ( x βˆ’ x β€² ) [ ( x βˆ’ x β€² ) 2 + ( y βˆ’ y β€² ) 2 + ( z βˆ’ z β€² ) 2 ] 3 / 2 x ^ \displaystyle \phantom{ \mathbf{\hat{x}}\frac{\partial g}{\partial x}} = -\frac{(x-x')}{\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{3/2}}\mathbf{\hat{x}}

    Other terms have the same denominator with variables and unit vector in numerator changed.

    y ^ βˆ‚ g βˆ‚ y = βˆ’ ( y βˆ’ y β€² ) [ ( x βˆ’ x β€² ) 2 + ( y βˆ’ y β€² ) 2 + ( z βˆ’ z β€² ) 2 ] 3 / 2 y ^ \displaystyle \mathbf{\hat{y}}\frac{\partial g}{\partial y} = -\frac{(y-y')}{\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{3/2}}\mathbf{\hat{y}}

    z ^ βˆ‚ g βˆ‚ z = βˆ’ ( z βˆ’ z β€² ) [ ( x βˆ’ x β€² ) 2 + ( y βˆ’ y β€² ) 2 + ( z βˆ’ z β€² ) 2 ] 3 / 2 z ^ \displaystyle \mathbf{\hat{z}}\frac{\partial g}{\partial z} = -\frac{(z-z')}{\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{3/2}}\mathbf{\hat{z}}

    The sum of the above three terms is an acceptable answer but note that it simplifies to

    βˆ‡ ( 1 ∣ r βˆ’ r β€² ∣ ) = βˆ’ ∣ r βˆ’ r β€² ∣ 3 ∣ r βˆ’ r β€² ∣ 3 = βˆ’ Τ‰ ^ 2 Τ‰ 2 \boldsymbol{\nabla}\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right)=\displaystyle -\frac{\phantom{|}\mathbf{r}-\mathbf{r}'\phantom{|^3}}{|\mathbf{r}-\mathbf{r}'|^3}=-\frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}

    because the denominator in each term is ∣ r βˆ’ r β€² ∣ 3 |\mathbf{r}-\mathbf{r}'|^3 . This can also be written as βˆ’ r ^ / r 2 -\mathbf{\hat{r}}/r^2

  2.  

    a. Substitute Τ‰ = ∣ r βˆ’ r β€² ∣ \char"0509=|\mathbf{r}-\mathbf{r}'| and Τ‰ ^ = ∣ r βˆ’ r β€² ∣ ∣ r βˆ’ r β€² ∣ \displaystyle {\hat{\textbf{\char"0509}}}=\frac{\phantom{|}\mathbf{r}-\mathbf{r}'\phantom{|}}{|\mathbf{r}-\mathbf{r}'|} into Τ‰ ^ 2 Τ‰ 2 \displaystyle \frac{\hat{\textbf{\char"0509}}\phantom{^2}}{\char"0509^2}

    b. Use r = x x ^ + y y ^ + z z ^ \mathbf{r}=x\mathbf{\hat{x}}+y\mathbf{\hat{y}}+z\mathbf{\hat{z}} and r β€² = x β€² x ^ + y β€² y ^ + z β€² z ^ \mathbf{r}'=x'\mathbf{\hat{x}}+y'\mathbf{\hat{y}}+z'\mathbf{\hat{z}} to get

    Τ‰ ^ Τ‰ 2 = ( x βˆ’ x β€² ) x ^ + ( y βˆ’ y β€² ) y ^ + ( z βˆ’ z β€² ) z ^ [ ( x βˆ’ x β€² ) 2 + ( y βˆ’ y β€² ) 2 + ( z βˆ’ z β€² ) 2 ] 3 / 2 \displaystyle \frac{\hat{\textbf{\char"0509}}}{\char"0509^2} = \frac{(x-x')\mathbf{\hat{x}}+(y-y')\mathbf{\hat{y}}+(z-z')\mathbf{\hat{z}}}{\left[(x-x')^2+(y-y')^2+(z-z')^2\right]^{3/2}}

2 Tangent Vectors

An object is moved along the path y = x 2 y=x^2 from x = 0 x=0 to x = x o x=x_o when there is an external force of F = βˆ’ m g y ^ \mathbf{F}=-mg\mathbf{\hat{y}} .

1. Compute the force tangent to the path at x o / 2 x_o/2 and x o x_o .

2. Compute the force perpendicular to the path at x o / 2 x_o/2 and x o x_o .

3. (Extra credit) Compute ∫ L F β‹… d l \int_{\mathcal{L}}\mathbf{F}\cdot d\mathbf{l} where L \mathcal{L} is the path along y = x 2 y=x^2 from x = 0 x=0 to x = x o x=x_o .

Answer

For 1. and 2., The problem statement does indicate if the answer should be a vector or scalar. I accepted either. (By convention, it is a vector.)

1.

t ^ = x ^ + 2 x y ^ 1 + 4 x 2 \displaystyle \mathbf{\hat{t}}=\frac{\mathbf{\hat{x}} + 2x\mathbf{\hat{y}}}{\sqrt{1+4x^2}}

F βˆ₯ = F β‹… t ^ = ( βˆ’ m g y ^ ) β‹… x ^ + 2 x y ^ 1 + 4 x 2 = βˆ’ 2 m g x 1 + 4 x 2 \displaystyle F_{\parallel} = \mathbf{F}\boldsymbol{\cdot} \mathbf{\hat{t}} = (-mg\mathbf{\hat{y}})\boldsymbol{\cdot} \frac{\mathbf{\hat{x}} + 2x\mathbf{\hat{y}}}{\sqrt{1+4x^2}}= -\frac{2mgx}{\sqrt{1+4x^2}}

This is a scalar b/c of the dot product. It is the component of force in the direction of chosen t ^ \mathbf{\hat{t}} . The negative is there because the component of F \mathbf{F} is in the opposite direction of the chosen t ^ \mathbf{\hat{t}} .

F βˆ₯ = F βˆ₯ t ^ = βˆ’ 2 m g x 1 + 4 x 2 x ^ + 2 x y ^ 1 + 4 x 2 = βˆ’ 2 m g x 1 + 4 x 2 ( x ^ + 2 x y ^ ) \displaystyle \mathbf{F}_{\parallel} = F_\parallel \mathbf{\hat{t}}= -\frac{2mgx}{\sqrt{1+4x^2}}\frac{\mathbf{\hat{x}} + 2x\mathbf{\hat{y}}}{\sqrt{1+4x^2}}=-\frac{2mgx}{1+4x^2}(\mathbf{\hat{x}} + 2x\mathbf{\hat{y}})

For x > 0 x>0 , the direction of this vector is down and to the left, as expected.

2.

n ^ = βˆ’ 2 x x ^ + y ^ 1 + 4 x 2 \displaystyle \mathbf{\hat{n}}=\frac{-2x\mathbf{\hat{x}}+\mathbf{\hat{y}}}{\sqrt{1+4x^2}}

F βŠ₯ = F β‹… n ^ = βˆ’ m g 1 + 4 x 2 \displaystyle F_{\perp} = \mathbf{F}\boldsymbol{\cdot} \mathbf{\hat{n}}=\frac{-mg}{\sqrt{1+4x^2}}

This is a scalar b/c of the dot product. It is the component of force in the direction of chosen n ^ \mathbf{\hat{n}} . The negative is there because the component of F \mathbf{F} component of F \mathbf{F} is the opposite direction of the chosen n ^ \mathbf{\hat{n}} .

F βŠ₯ = F βŠ₯ n ^ = m g 1 + 4 x 2 ( 2 x x ^ βˆ’ y ^ ) \displaystyle \mathbf{F}_{\perp} = F_\perp \mathbf{\hat{n}}=\frac{mg}{1+4x^2}(2x\mathbf{\hat{x}}-\mathbf{\hat{y}})

For x > 0 x>0 , the direction of this vector is to the right and down, as expected.

Could also compute F βŠ₯ \mathbf{F}_\perp using F βŠ₯ = F βˆ’ F βˆ₯ \mathbf{F}_\perp = \mathbf{F}-\mathbf{F}_\parallel .

3. The problem asks to integrate along the line. For the path given, d l = d x 1 + 2 x 2 dl = dx\sqrt{1+2x^2} and d l = t ^ d l d\mathbf{l}=\mathbf{\hat{t}}dl , so F β‹… d l = βˆ’ m g d x \mathbf{F}\boldsymbol{\cdot} d\mathbf{l} = -mgdx . The integral is then ∫ 0 x o ( βˆ’ 2 m g x d x ) = βˆ’ m g x o 2 \int_{0}^{x_o}(-2mgxdx)=-mgx_o^2 . The integral corresponds to the work done moving the object a height h = x o 2 h=x_o^2 . The easier way to solve this is to to note that F \mathbf{F} is a conservative force and simply write m g h mgh and then plug in h = x o 2 h=x_o^2 . (Ideally I would have given the equation as y = x 2 / x o y=x^2/x_o so h h would not look like it is the square of a distance. Technically the problem statement is still correct because impicitly x o x_o is dimensionless.)1.2.3.

3 Normal Vectors

A plane has corners at ( x , y , z ) = ( 0 , 0 , 0 ) (x,y,z)=(0,0,0) , ( x , y , z ) = ( 1 , 0 , 0 ) (x,y,z)=(1,0,0) , ( x , y , z ) = ( 0 , 1 , 1 ) (x,y,z)=(0,1,1) , and ( x , y , z ) = ( 1 , 1 , 1 ) (x,y,z)=(1,1,1) .

1. Sketch the plane with a 3–dimensional diagram.

2. Sketch the plane in the y y – z z plane. (That is, what you would see if you looked toward the origin from a point with large x x .)

3. Find a vector normal to the plane that has a positive z z –component.

4. There is a force acting on the plane of F = F x x ^ + F y y ^ + F z z ^ \mathbf{F}=F_x\mathbf{\hat{x}}+F_y\mathbf{\hat{y}}+F_z\mathbf{\hat{z}} . Find the component of force perpendicular to the plane.

Partial Answer

3. n ^ = ( βˆ’ y ^ + z ^ ) / 2 \mathbf{\hat{n}}=(-\mathbf{\hat{y}} + \mathbf{\hat{z}})/\sqrt{2} ; this can be computed using the diagram or by using the u Γ— v \mathbf{u}\times \mathbf{v} method.

4. F βŠ₯ = F β‹… n ^ = βˆ’ F y + F z F_\perp = \mathbf{F}\boldsymbol{\cdot} \mathbf{\hat{n}}=-F_y+F_z and F βŠ₯ = F βŠ₯ n ^ = ( βˆ’ F y + F z ) ( βˆ’ y ^ + z ^ ) / 2 \mathbf{F}_\perp=F_\perp\mathbf{\hat{n}}=(-F_y+F_z)(-\mathbf{\hat{y}} + \mathbf{\hat{z}})/\sqrt{2} . I accepted either answer because the question does not specify if it wants a scalar or vector for the force. (By convention, it is a vector.). Ideally you drew F \mathbf{F} and n ^ \mathbf{\hat{n}} on the diagram for 2. to make sure that the direction of F βŠ₯ \mathbf{F}_\perp made sense by plugging in, say, F y = F z = 1 F_y=F_z=1 and F y = 0 F_y=0 and F z = 1 F_z=1 .

A common error was to write F β‹… n ^ = βˆ’ F y y ^ + F z z ^ \mathbf{F}\boldsymbol{\cdot} \mathbf{\hat{n}}=-F_y\mathbf{\hat{y}}+F_z\mathbf{\hat{z}} . This does not make sense because a dot product should result in a scalar.

4 Notation and Vector Fields

A positive charge q q is placed at ( x , y ) = ( x o , 0 ) (x,y)=(x_o,0) and a negative charge βˆ’ q -q at ( x , y ) = ( βˆ’ x o , 0 ) (x,y)=(-x_o,0) .

The force on a charge Q Q due to q q and βˆ’ q -q is

F = k q Q Τ‰ ^ + Τ‰ + 2 βˆ’ k q Q Τ‰ ^ βˆ’ Τ‰ βˆ’ 2 \mathbf{F}=kqQ\frac{\hat{\textbf{\char"0509}}_+}{\char"0509^2_+}-kqQ\frac{\hat{\textbf{\char"0509}}_{-}}{\char"0509^2_{-}}

where Τ‰ + = r βˆ’ r + \textbf{\char"0509}_+=\mathbf{r}-\mathbf{r_+} , Τ‰ βˆ’ = r βˆ’ r βˆ’ \textbf{\char"0509}_-=\mathbf{r}-\mathbf{r_-} , r = x x ^ + y y ^ \mathbf{r}=x\mathbf{\hat{x}} + y\mathbf{\hat{y}} , r + \mathbf{r_+} is the vector from the origin to the positive charge, and r βˆ’ \mathbf{r_-} is the vector from the origin to the negative charge.

Find F \mathbf{F} in cartesian coordinates with cartesian unit vectors in terms of the constants given at

1. ( x , y ) = ( 2 , 0 ) (x,y)=(2,0)

2. ( x , y ) = ( 0 , 2 ) (x,y)=(0,2)

It may help to solve this by using the techniques from Physics 260 first. This is a straightforward problem that is written in the notation used by Griffiths.

Answer

F ( x , y ) = k q Q [ ( x βˆ’ x o ) x ^ + y y ^ [ ( x βˆ’ x o ) 2 + y 2 ] 3 / 2 βˆ’ ( x + x o ) x ^ + y y ^ [ ( x + x o ) 2 + y 2 ] 3 / 2 ] \displaystyle\mathbf{F}(x,y) = kqQ\left[\frac{(x-x_o)\mathbf{\hat{x}}+y\mathbf{\hat{y}}}{\left[(x-x_o)^2+y^2\right]^{3/2}}-\frac {(x+x_o)\mathbf{\hat{x}}+y\mathbf{\hat{y}}}{\left[(x+x_o)^2+y^2\right]^{3/2}}\right]

1. F = k q Q [ 2 βˆ’ x o ∣ 2 βˆ’ x o ∣ 3 βˆ’ 2 + x o ∣ 2 + x o ∣ 3 ] x ^ \displaystyle\mathbf{F} = kqQ\left[\frac{2-x_o}{|2-x_o|^3}-\frac {2+x_o}{|2+x_o|^3}\right]\mathbf{\hat{x}} (In an earlier version of the solutions, had simplified this as k q Q [ 1 ( 2 βˆ’ x o ) 2 βˆ’ 1 ( 2 + x o ) 2 ] x ^ kqQ\left[\frac{1}{(2-x_o)^2}-\frac{1}{(2+x_o)^2}\right]\mathbf{\hat{x}} , but this simplification is only applicable for x o < 2 x_o \lt 2 ; the problem statement should have had the positions to find the field at of ( 2 x o , 0 ) (2x_o,0) and ( 0 , 2 x o ) (0, 2x_o) , in which case the cancellation is allowed).

2. F = k q Q [ βˆ’ x o x ^ + 2 y ^ [ x o 2 + 2 2 ] 3 / 2 βˆ’ x o x ^ + 2 y ^ [ x o 2 + 2 2 ] 3 / 2 ] = βˆ’ k q Q 2 x o [ 1 [ x o 2 + 2 2 ] 3 / 2 ] x ^ \displaystyle\mathbf{F} = kqQ\left[\frac{-x_o\mathbf{\hat{x}}+2\mathbf{\hat{y}}}{\left[x_o^2+2^2\right]^{3/2}}-\frac {x_o\mathbf{\hat{x}}+2\mathbf{\hat{y}}}{\left[x_o^2+2^2\right]^{3/2}}\right] = -kqQ2x_o\left[\frac{1}{\left[x_o^2+2^2\right]^{3/2}}\right]\mathbf{\hat{x}}

5 Field Lines and Equipotentials

1. Draw field lines associated with the two starting points shown in the diagram for A = βˆ’ sin ⁑ Ο• x ^ + cos ⁑ Ο• y ^ \mathbf{A}=-\sin\phi\mathbf{\hat{x}} + \cos\phi\mathbf{\hat{y}} . Draw the field lines until they encounter the βˆ’ x -x –axis.

2. Sketch four equipotential lines.

1. Field lines are circles.

2. β€œEquipotentials” are radial lines that intersect the origin. As noted in class, this is not a conservative field and so it does not make sense to draw equipotentials. I meant to ask you to draw lines that are always perpendicular to the field lines. This vector field is not conservative because the integral ∫ L A β‹… d l \int_{\mathcal{L}}\mathbf{A}\boldsymbol{\cdot} d\mathbf{l} is not the same for all paths that have the same starting and ending points. For example, if L 1 \mathcal{L}_1 is a single circle with a radius of 1 1 , the integral is 2 Ο€ 2\pi . If L 2 \mathcal{L}_2 is L 1 \mathcal{L}_1 repeated twice, the integral is 4 Ο€ 4\pi .