1. Vector Fields
    1. Example – Vector Field with Cartesian Coordinates and Unit Vectors
    2. Example – Radial Field With Cylindrical Coordinate and Unit Vector
    3. Problem – Radial Field in Spherical
    4. Example – Fields with Cartesian Unit Vectors and Cylindrical Coordinates
    5. Problem – Fields with Cartesian Unit Vectors and Cylindrical Coordinates
    6. Problem – Fields with Cartesian Unit Vectors and Cartesian Coordinates
    7. Problem – Fields with Cylindrical Unit Vectors and Cylindrical Coordinates

If you have difficulty solving these problems, additional background material is available at Vector Fields.

You can experiment with the creation of vector fields using a GeoGebra app.

A vector field diagram is a plot showing the vectors associated with a vector function at select points in space.

Vector fields are fundamental to the theory of E&M. Many of the fundamental theorems involve derivatives and integrals of vector functions that depend on position.

An example of a vector function is the electric field due to a positive point charge Q Q at the origin:

E ( r ) = k Q r 2 r ^ \mathbf{E}(r)=\frac{kQ}{r^2}\boldsymbol{\hat{r}}

The vector field diagram of A ≑ E ( r ) / k Q = r ^ / r 2 \mathbf{A}\equiv\mathbf{E}(r)/kQ = \boldsymbol{\hat{r}}/r^2 is

Generally, only the relative lengths of the vectors are of interest and so a scale allowing one to determine a numeric value of the magnitude of the vectors was omitted in this figure. Because the outer vectors are at a position that is 4x larger than the inner vectors, the outer vectors have a length that is 1/16 of the inner vectors.

When drawing a vector field diagram, points must be selected where a vector is drawn. The general rule is that just enough points that a reader can determine the general length and direction of vectors at points without a vector on the diagram. In the previous figure, it is clear that the vectors are radial, the magnitude is is independent of radius, and the magnitude is proportional to 1 / r 2 1/r^2 . These three facts are the key patterns associated with the vector function plotted.

Plot the vector field A = x x ^ + y y ^ \mathbf{A} = x\hat{\mathbf{x}} + y\hat{\mathbf{y}} at x = 1 x=1 and y = βˆ’ 1 , 0 , 1 y=-1, 0, 1 .

Answer: At the given points, A = ∣ A ∣ = 2 , 1 , 2 A=|\mathbf{A}|=\sqrt{2}, 1, \sqrt{2} . In the following diagram, the grid spacing has been assumed to be 1 1 . Note that we could have also solved this probelem in cylindrical coordinates using A = s s ^ \mathbf{A} = s\mathbf{\hat{s}} . In this case, the values of s s would be given by the magnitudes of A A already computed and we would need to determine the angles of the vector at each point. For example, the angle at x = 1 x=1 , y = βˆ’ 1 y=-1 would be tan ⁑ βˆ’ 1 ( + 1 βˆ’ 1 ) = 31 5 ∘ \tan^{-1}\left(\frac{+1}{-1}\right) = 315^\circ .

Outside of a solid and long cylinder of radius R R with a uniform linear charge density of Ξ» \lambda , the field is

E ( r ) = 2 k Ξ» s ^ s \mathbf{E}(r)=2k\lambda\frac{\boldsymbol{\hat{s}}}{s}

inside, it is

E ( r ) = 2 k Ξ» s R 2 s ^ \mathbf{E}(r)=2k\lambda\frac{s}{R^2}\boldsymbol{\hat{s}}

Sketch the vector field A ≑ E / ( 2 k Ξ» ) \mathbf{A}\equiv\mathbf{E}/(2k\lambda) inside and outside of the cylinder if the cylinder is along the z z –axis and centered on the origin as shown in the following diagram.

Answer: Assume that R = 1 R = 1 . As noted, only the relative lengths matter in a vector field diagram. Points at s = R / 2 s=R/2 are inside the cylinder and so the inside equation is used, giving A = s ^ / 2 \mathbf{A}=\boldsymbol{\hat{s}}/2 . At s = R s=R , we can use either formula – they give the same result of A = s ^ \mathbf{A}=\boldsymbol{\hat{s}} . At s = 1.5 R s=1.5R , A = 2 s ^ / 3 \mathbf{A}=2\boldsymbol{\hat{s}}/3 . At s = 2 R s=2R , A = s ^ / 2 \mathbf{A}=\boldsymbol{\hat{s}}/2 . As one moves from the origin outwards, the vectors increase in length until s = R s=R at which point their length decreases as s s increases. At a given Ο• \phi , the vectors are all the same length.

Outside of a solid sphere of radius R R with uniformly distributed charge Q Q , the field is

E ( s ) = k Q 1 r 2 r ^ \mathbf{E}(s)=kQ\frac{1}{r^2}\boldsymbol{\hat{r}}

inside, it is

E ( s ) = k Q r R 3 r ^ \mathbf{E}(s)=kQ\frac{r}{R^3}\boldsymbol{\hat{r}}

Sketch the vector field A ≑ E / ( k Q ) \mathbf{A}\equiv\mathbf{E}/(kQ) inside and outside of the sphere.

Plot the vector A = cos ⁑ Ο• x ^ + sin ⁑ Ο• y ^ \mathbf{A}=\cos\phi\hat{\mathbf{x}} + \sin\phi\hat{\mathbf{y}} at the points shown in the following diagram.

Plot the vector A = βˆ’ sin ⁑ Ο• x ^ + cos ⁑ Ο• y ^ \mathbf{A}=-\sin\phi\hat{\mathbf{x}} + \cos\phi\hat{\mathbf{y}} at the points shown in the following diagram.

Plot the vector A = x s x ^ + y s y ^ \displaystyle\mathbf{A}=\frac{x}{s}\hat{\mathbf{x}} + \frac{y}{s}\hat{\mathbf{y}} at the points shown in the following diagram. Recall that the radial coordinate in cylindrical coordinates is s s and s = x 2 + y 2 s=\sqrt{x^2+y^2} .

Plot the vector A = βˆ’ y s x ^ + x s y ^ \displaystyle\mathbf{A}=-\frac{y}{s}\hat{\mathbf{x}} + \frac{x}{s}\hat{\mathbf{y}} at the points shown in the following diagram.

Plot the vector A = s Ο• ^ \mathbf{A}=s\hat{\boldsymbol{\phi}} at the points shown in the following diagram.