The electric field

Electric field generated by a point charge

The concept of an electric field was introduced to suggest how electric charges could exert forces even when they were not touching each other. That could not be explained unless a sort of mediator “carried” the force from a charge to another, hence the idea of field: charges do not directly exert a force - they produce an electric field, which then in turn exerts a force on other charges.

Although this may seem just a mathematical trick, we will see later on that the electric field actually exists and propagates at a finite speed.

Let's consider a point charge $Q$$Q$ at the origin of an orthogonal coordinate system and let's imagine to place a test charge $q_{0}$$q_0$ at a point ${\underline {r}}$$\underline{r}$ ($q_{0}$$q_0$ is so small that any effects due to its presence are considered negligible). The Coulomb's force, due to the presence of $Q$$Q$, acting on the test charge is: ${\underline {F}}={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Qq_{0}}{r^{2}}}{\underline {u_{r}}}\ .$${\underline {F}}={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Qq_{0}}{r^{2}}}{\underline {u_{r}}}\ .$

We then define the electric field produced by $Q$$Q$ at the point ${\underline {r}}$$\underline{r}$ as: $E_{Q}\left({\underline {r}}\right)={\frac {F}{q_{0}}}={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q}{r^{2}}}{\underline {u_{r}}}\ .$$E_{Q}\left({\underline {r}}\right)={\frac {F}{q_{0}}}={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q}{r^{2}}}{\underline {u_{r}}}\ .$

From this definition, we can say that if an electric field ${\underline {E}}$$\underline{E}$ is present in a given point, then a charge $q$$q$ in that same point will experience the following force: ${\underline {F}}=q{\underline {E}}\ .$${\underline {F}}=q{\underline {E}}\ .$

Field lines

Fields lines give a graphical representation of how the electric field interacts with the surrounding space.

They have some key features:

• They are tangent to the field at any point;
• They have a direction;
• By convention they start from a positive charge and end to a negative one;
• Any two field lines never intersect;
• Qualitatively, the more dense the field lines, the more intense the electric field.

The principle of superposition is valid for the field too, therefore when there are multiple charges, the field generated by each of them interacts with the field of the other charges.

In particular, the field generated by two charges of opposite sign will have the field lines arranged as follows (such a structure is called electric dipole):

whereas two charges of same sign will produce an arrangement of the field lines similar to the following diagram:

Superposition principle

Imagine we have a system of $n$$n$ point charges $q_{i}$$q_i$, each placed at a point $r_{i}$$r_i$. If a charge $Q$$Q$ is located at the point $r$$r$, you can calculate the force acting on it by adding vectorially the forces exerted by each individual charge. The same thing can be done with the electric field: the electric field generated at $r$$r$ by the system of point charges is

$${\underline {E}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{n}{\frac {q_{i}\left({\underline {r}}-{\underline {r_{i}}}\right)}{\left|{\underline {r}}-{\underline {r_{i}}}\right|^{3}}}\ .$$

$$\underline{E}= \frac {1}{4 \pi \varepsilon_0} \sum_{i=1}^n {\frac {q_i\left(\underline{r} - \underline{r_i}\right)}{\left|\underline{r} - \underline{r_i}\right|^3}} \ .$$

Continuous distributions of charge

The superposition principle can also be applied to continuous distributions of charge.

Consider a continuous distribution of charge, arranged along a curve $\Gamma$$\Gamma$ with a linear charge density $\lambda$$\lambda$. A section $dr'$$dr'$ contains a charge $dq=\lambda dr'$$dq = \lambda dr'$. Denoting by ${\underline {r'}}$$\underline{r'}$ the points of the charge distribution, the field generated by the distribution at the point ${\underline {r}}$$\underline{r}$ is given by the integral:

$${\underline {E_{\Gamma }}}\left({\underline {r}}\right)={\frac {1}{4\pi \varepsilon _{0}}}\int _{\Gamma }{\frac {\lambda \left({\underline {r}}-{\underline {r'}}\right)}{\left|{\underline {r}}-{\underline {r'}}\right|^{3}}}d{\underline {r'}}\ .$$

$$\underline{E_{\Gamma}}\left(\underline{r}\right) = \frac{1}{4 \pi \varepsilon_0} \int_{\Gamma} \frac{\lambda\left(\underline{r}-\underline{r'}\right)}{\left| \underline{r} - \underline{r'} \right| ^3} d\underline{r'} \ .$$

The same reasoning can be applied to surface charge distributions $\sigma$$\sigma$ or volumetric ones $\rho$$\rho$. The results are respectively:

$${\underline {E_{\Sigma }}}\left({\underline {r}}\right)={\frac {1}{4\pi \varepsilon _{0}}}\int _{\Sigma }{\frac {\sigma \left({\underline {r}}-{\underline {r'}}\right)}{\left|{\underline {r}}-{\underline {r'}}\right|^{3}}}d\Sigma '\ ,$$

$$\underline{E_{\Sigma}}\left(\underline{r}\right) = \frac{1}{4 \pi \varepsilon_0} \int_{\Sigma} \frac{\sigma\left(\underline{r}-\underline{r'}\right)}{\left| \underline{r} - \underline{r'} \right| ^3} d\Sigma' \ ,$$

$${\underline {E_{V}}}\left({\underline {r}}\right)={\frac {1}{4\pi \varepsilon _{0}}}\int _{V}{\frac {\rho \left({\underline {r}}-{\underline {r'}}\right)}{\left|{\underline {r}}-{\underline {r'}}\right|^{3}}}dV'\ .$$

$$\underline{E_{V}}\left(\underline{r}\right) = \frac{1}{4 \pi \varepsilon_0} \int_{V} \frac{\rho\left(\underline{r}-\underline{r'}\right)}{\left| \underline{r} - \underline{r'} \right| ^3} dV' \ .$$

Electric field generated by a uniformly charged wire

Let's consider a uniformly charged wire of length $2L$$2L$, with a linear charge density $\lambda$$\lambda$, and let's calculate the electric field generated by this distribution of charge at a general point in space.

First, let's fix an orthogonal coordinate system: the origin is set at the centre of the wire, and the distribution of charge is placed along the $z$$z$ axis.

Let's then denote by ${\underline {r}}$$\underline{r}$ a generic point of the space at which we want to calculate the electric field, and ${\underline {r}}'$$\underline{r}'$ a point of the distribution of charge. We will have:

$${\underline {r}}=\left(x,y,z\right)\ ,\qquad {\underline {r}}'=\left(0,0,z'\right)\ .$$

$$\underline{r} = \left(x,y,z\right) \ , \qquad \underline{r}' = \left(0,0,z'\right) \ .$$

Hence:

$${\underline {r}}-{\underline {r}}'=\left(x,y,z-z'\right)\ ,$$

$$\underline{r} - \underline{r}' = \left(x,y,z-z'\right) \ ,$$

$$\left|{\underline {r}}-{\underline {r}}'\right|^{3}=\left[x^{2}+y^{2}+\left(z-z'\right)^{2}\right]^{\frac {3}{2}}\ .$$

$$\left| \underline{r} - \underline{r}' \right|^3 = \left[x^2+y^2+\left(z-z'\right)^2\right]^ {\frac{3}{2}} \ .$$

We can then proceed to the calculation of the electric field by calculating the following integral:

$${\underline {E}}\left(x,y,z\right)={\frac {\lambda }{4\pi \varepsilon _{0}}}\int _{\Gamma }{\frac {{\underline {r}}-{\underline {r}}'}{\left|{\underline {r}}-{\underline {r}}'\right|^{3}}}dl\ .$$

$$\underline{E}\left(x,y,z\right) = \frac{\lambda}{4 \pi \varepsilon_0} \int_{\Gamma} \frac{\underline{r} - \underline{r}'}{\left| \underline{r} - \underline{r}' \right|^3} dl \ .$$

Now, substituting and taking into account that $dl=dz'$$dl=dz'$, the integral becomes:

$${\underline {E}}\left(x,y,z\right)={\frac {\lambda }{4\pi \varepsilon _{0}}}\int _{-L}^{L}{\frac {x\mathbf {e_{x}} +y\mathbf {e_{y}} +(z-z')\mathbf {e_{z}} }{\left[x^{2}+y^{2}+\left(z-z'\right)^{2}\right]^{\frac {3}{2}}}}dz'\ ,$$

$$\underline{E}\left(x,y,z\right) = \frac{\lambda}{4\pi\varepsilon_0} \int_{-L}^{L} \frac{x\mathbf{e_x}+y\mathbf{e_y}+(z-z')\mathbf{e_z}}{\left[ x^2 + y^2 +\left(z-z'\right)^2\right]^{\frac{3}{2}}} dz' \ ,$$

where $\mathbf {e_{x}}$$\mathbf{e_x}$, $\mathbf {e_{y}}$$\mathbf{e_y}$, $\mathbf {e_{z}}$$\mathbf{e_z}$ are the unit vectors of the Cartesian axes.

Now let's solve separately the integrals for the three components of the electric field. Let's start by calculating $E_{x}$$E_x$:

$$E_{x}={\frac {\lambda }{4\pi \varepsilon _{0}}}\int _{-L}^{L}{\frac {x}{\left[x^{2}+y^{2}+\left(z-z'\right)^{2}\right]^{\frac {3}{2}}}}dz'\ .$$

$$E_x = \frac{\lambda}{4 \pi \varepsilon_0} \int_{-L}^{L} \frac{x}{\left[ x^2 + y^2 +\left(z-z'\right)^2\right]^{\frac{3}{2}}} dz' \ .$$

We carry out the following substitution:

$$z-z'={\sqrt {x^{2}+y^{2}}}\sinh t\quad \Rightarrow \quad dz'=-{\sqrt {x^{2}+y^{2}}}\cosh t\,dt\ .$$

$$z-z'=\sqrt{x^2+y^2}\sinh t \quad \Rightarrow \quad dz'=-\sqrt{x^2+y^2}\cosh t \,dt \ .$$

And now the indefinite integral can be solved:

$$\int {\frac {x{\sqrt {x^{2}+y^{2}}}\cosh t}{\left[\left(x^{2}+y^{2}\right)+\left(x^{2}+y^{2}\right)\sinh ^{2}t\right]^{\frac {3}{2}}}}dt={\frac {x}{x^{2}+y^{2}}}\int {\frac {1}{\cosh ^{2}t}}dt={\frac {x}{x^{2}+y^{2}}}\tanh t\ .$$

$$\int \frac{x\sqrt{x^2+y^2} \cosh t}{\left[\left(x^2+y^2\right)+\left(x^2+y^2\right)\sinh^2 t\right]^{\frac{3}{2}}} dt = \frac{x}{x^2+y^2}\int\frac{1}{\cosh^2t}dt= \frac{x}{x^2+y^2} \tanh t \ .$$

Taking into account the fact that, from our substitution:

$$\sinh t={\frac {z-z'}{\sqrt {x^{2}+y^{2}}}}\ ,$$

$$\sinh t = \frac{z-z'}{\sqrt{x^2+y^2}} \ ,$$

and using the relationship:

$$\tanh t={\frac {\sinh t}{\sqrt {1+\sinh ^{2}t}}}\ ,$$

$$\tanh t = \frac{\sinh t}{\sqrt{1+\sinh^2 t}} \ ,$$

we can proceed to the calculation of the definite integral:

$$E_{x}=-{\frac {\lambda }{4\pi \varepsilon _{0}}}\left({\frac {x}{x^{2}+y^{2}}}{\frac {\left(z-z'\right)}{\sqrt {x^{2}+y^{2}+\left(z-z'\right)^{2}}}}\right){\Bigg |}_{-L}^{L}\ ,$$

$$E_x= -\frac{\lambda}{4\pi\varepsilon_0} \left( \frac{x}{x^2+y^2} \frac{\left(z-z'\right)}{\sqrt{x^2+y^2+\left(z-z'\right)^2}} \right) \Bigg|_{-L}^{L} \ ,$$

$$E_{x}={\frac {\lambda }{4\pi \varepsilon _{0}}}{\frac {x}{x^{2}+y^{2}}}\left({\frac {z+L}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}-{\frac {z-L}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}\right)\ .$$

$$E_x= \frac{\lambda}{4\pi\varepsilon_0}\frac{x}{x^2+y^2} \left( \frac{z+L}{\sqrt{x^2+y^2+\left(z+L\right)^2}} - \frac{z-L}{\sqrt{x^2+y^2+\left(z-L\right)^2}} \right) \ .$$

The calculation of the component $E_{y}$$E_y$ is analogous, it leads to the following result:

$$E_{y}={\frac {\lambda }{4\pi \varepsilon _{0}}}{\frac {y}{x^{2}+y^{2}}}\left({\frac {z+L}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}-{\frac {z-L}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}\right)\ .$$

$$E_y= \frac{\lambda}{4\pi\varepsilon_0}\frac{y}{x^2+y^2} \left( \frac{z+L}{\sqrt{x^2+y^2+\left(z+L\right)^2}} - \frac{z-L}{\sqrt{x^2+y^2+\left(z-L\right)^2}} \right) \ .$$

We can now proceed with the calculation of the component $E_{z}$$E_z$:

$$E_{z}={\frac {\lambda }{4\pi \varepsilon _{0}}}\int _{-L}^{L}{\frac {z-z'}{\left[x^{2}+y^{2}+\left(z-z'\right)^{2}\right]^{\frac {3}{2}}}}dz'\ .$$

$$E_z = \frac{\lambda}{4\pi\varepsilon_0}\int_{-L}^{L} \frac{z-z'}{\left[ x^2 + y^2 +\left(z-z'\right)^2\right]^{\frac{3}{2}}} dz' \ .$$

This time we carry out a different substitution:

$$x^{2}+y^{2}+\left(z-z'\right)^{2}=t\quad \Rightarrow \quad -2\left(z-z'\right)dz'=dt\ .$$

$$x^2 + y^2 +\left(z-z'\right)^2 = t \quad\Rightarrow\quad -2\left(z-z'\right)dz'=dt \ .$$

Hence:

$$E_{z}={\frac {-\lambda }{4\pi \varepsilon _{0}}}{\frac {1}{2}}\int _{a}^{b}{\frac {dt}{t^{\frac {3}{2}}}}\ .$$

$$E_z = \frac{-\lambda}{4\pi\varepsilon_0} \frac{1}{2}\int_{a}^{b} \frac{dt}{t^{\frac{3}{2}}} \ .$$

Using $a=x^{2}+y^{2}+\left(z+L\right)^{2}$$a=x^2 + y^2 +\left(z+L\right)^2$ e $b=x^{2}+y^{2}+\left(z-L\right)^{2}$$b=x^2 + y^2 +\left(z-L\right)^2$. and integrating, we get:

$$E_{z}={\frac {\lambda }{4\pi \varepsilon _{0}}}\left({\frac {1}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}-{\frac {1}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}\right)\ .$$

$$E_z = \frac{\lambda}{4\pi\varepsilon_0} \left( \frac{1}{\sqrt{x^2+y^2+\left(z-L\right)^2}}-\frac{1}{\sqrt{x^2+y^2+\left(z+L\right)^2}} \right) \ .$$

Combining everything that has been calculated so far, the final result is:

$${\underline {E}}\left(x,y,z\right)={\frac {\lambda }{4\pi \varepsilon _{0}}}{\begin{pmatrix}{\frac {x}{x^{2}+y^{2}}}\left({\frac {z+L}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}-{\frac {z-L}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}\right)\\{\frac {y}{x^{2}+y^{2}}}\left({\frac {z+L}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}-{\frac {z-L}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}\right)\\{\frac {1}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}-{\frac {1}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}\end{pmatrix}}\ .$$

$${\underline {E}}\left(x,y,z\right)={\frac {\lambda }{4\pi \varepsilon _{0}}}{\begin{pmatrix}{\frac {x}{x^{2}+y^{2}}}\left({\frac {z+L}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}-{\frac {z-L}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}\right)\\{\frac {y}{x^{2}+y^{2}}}\left({\frac {z+L}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}-{\frac {z-L}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}\right)\\{\frac {1}{\sqrt {x^{2}+y^{2}+\left(z-L\right)^{2}}}}-{\frac {1}{\sqrt {x^{2}+y^{2}+\left(z+L\right)^{2}}}}\end{pmatrix}}\ .$$

Electric field generated by a uniformly charged disk

Let's consider a uniformly charged disc of radius $R$$R$, with surface charge density $\sigma$$\sigma$, and let's calculate the electric field generated along its $z$$z$ axis. A cylindrical coordinate system is set such that the disk is in the $xy$$xy$ plane, and its centre is at the origin.

A generic point of the axis of the disc can be expressed in the form:

$${\underline {r}}=\left(0,0,z\right),$$

$$\underline{r}=\left(0,0,z\right) ,$$

and a generic point of the disc will be:

$${\underline {r}}'=\left(\rho \cos \theta ,\rho \sin \theta ,0\right)\ .$$

$$\underline{r}'= \left(\rho \cos\theta, \rho \sin \theta, 0 \right) \ .$$

We therefore obtain:

$${\underline {r}}-{\underline {r}}'=\left(-\rho \cos \theta ,-\rho \sin \theta ,z\right)\ ,\qquad \quad \left|{\underline {r}}-{\underline {r}}'\right|^{3}=\left(\rho ^{2}+z^{2}\right)^{\frac {3}{2}}\ .$$

$$\underline{r}-\underline{r}'=\left(-\rho \cos\theta, -\rho \sin \theta, z \right) \ , \qquad\quad \left|\underline{r}-\underline{r}'\right|^3= \left( \rho^2+z^2 \right)^{\frac{3}{2}} \ .$$

By using the expressions above, it is possible to write the integral for calculating the electric field:

$${\underline {E}}\left(z\right)={\frac {\sigma }{4\pi \varepsilon _{0}}}\int _{D}{\frac {-\rho \cos \theta \mathbf {e_{x}} -\rho \sin \theta \mathbf {e_{y}} +z\mathbf {e_{z}} }{\left(\rho ^{2}+z^{2}\right)^{\frac {3}{2}}}}ds\ .$$

$$\underline{E}\left(z\right)= \frac{\sigma}{4\pi\varepsilon_0}\int_D \frac{-\rho \cos\theta\mathbf{e_x} -\rho \sin \theta\mathbf{e_y} + z\mathbf{e_z}}{\left( \rho^2+z^2 \right)^{\frac{3}{2}}} ds \ .$$

In cylindrical coordinates we have $ds=\rho d\theta d\rho$$ds=\rho d\theta d\rho$ with $\rho \in \left[0,R\right]$$\rho \in \left[0,R\right]$, $\theta \in \left[0,2\pi \right)$$\theta \in \left[0,2\pi\right)$, so the integral becomes:

$${\underline {E}}\left(z\right)={\frac {\sigma }{4\pi \varepsilon _{0}}}\int _{0}^{R}\left(\int _{0}^{2\pi }{\frac {-\rho \cos \theta \mathbf {e_{x}} -\rho \sin \theta \mathbf {e_{y}} +z\mathbf {e_{z}} }{\left(\rho ^{2}+z^{2}\right)^{\frac {3}{2}}}}\rho d\theta \right)d\rho \ .$$

$$\underline{E}\left(z\right)= \frac{\sigma}{4\pi\varepsilon_0}\int_{0}^{R}\left(\int_{0}^{2\pi} \frac{-\rho \cos\theta\mathbf{e_x} -\rho \sin \theta\mathbf{e_y} + z\mathbf{e_z}}{\left( \rho^2+z^2 \right)^{\frac{3}{2}}}\rho d\theta \right) d\rho \ .$$

The integral in $\theta$$\theta$ cancels out the contributions in $\mathbf {e_{x}}$$\mathbf{e_x}$ and $\mathbf {e_{y}}$$\mathbf{e_y}$, therefore just the integral of the field directed along the $z$$z$ axis is left:

$${\underline {E}}\left(z\right)={\frac {\sigma }{4\pi \varepsilon _{0}}}2\pi \int _{0}^{R}{\frac {z\rho \mathbf {e_{z}} }{\left(\rho ^{2}+z^{2}\right)^{\frac {3}{2}}}}d\rho \ .$$

$$\underline{E}\left(z\right)= \frac{\sigma}{4\pi\varepsilon_0} 2\pi \int_{0}^{R} \frac{z\rho\mathbf{e_z}}{\left( \rho^2+z^2 \right)^{\frac{3}{2}}}d\rho \ .$$

We use the substitution $\rho ^{2}+z^{2}=t\Rightarrow 2\rho d\rho =dt$$\rho^2 + z^2 = t \Rightarrow 2\rho d\rho= dt$ and we obtain:

$${\underline {E}}\left(z\right)={\frac {\sigma z\mathbf {e_{z}} }{4\varepsilon _{0}}}\int _{z^{2}}^{R^{2}+z^{2}}{\frac {dt}{t^{\frac {3}{2}}}}={\frac {\sigma z\mathbf {e_{z}} }{2\varepsilon _{0}}}\left({\frac {-1}{\sqrt {t}}}\right){\Bigg |}_{z^{2}}^{R^{2}+z^{2}}\ .$$

$$\underline{E}\left(z\right)= \frac{\sigma z \mathbf{e_z}}{4\varepsilon_0} \int_{z^2}^{R^2+z^2} \frac{dt}{t^{\frac{3}{2}}} = \frac{\sigma z \mathbf{e_z}}{2\varepsilon_0} \left( \frac{-1}{\sqrt{t}} \right) \Bigg|_{z^2}^{R^2+z^2} \ .$$

The calculation gives the final result:

$${\underline {E}}\left(z\right)={\frac {\sigma z\mathbf {e_{z}} }{2\varepsilon _{0}}}\left({\frac {1}{\left|z\right|}}-{\frac {1}{\sqrt {R^{2}+z^{2}}}}\right)\ .$$

$$\underline{E}\left(z\right) = \frac{\sigma z \mathbf{e_z}}{2\varepsilon_0}\left( \frac{1}{\left|z\right|}-\frac{1}{\sqrt{R^2+z^2}} \right) \ .$$