Electrostatic Force, Electric Fields and Potential

Electrical Forces

We have seen that masses attract each other through the gravitational force. Electrical charges also

Like charges repell each other. So two positive, or two negative charges brought together will push each other apart but two charges of different sign will create an attractive force. The magnitude of the force is given by

Columbs law states that when two charges are brought together, there is a force on each one. The force is proportional to the magnitude of their charges and inversely proportion to the square of the distance between them.

$F = -\frac{k q_1 q_2}{r^2}$

Where $k = \frac{4\pi}{\epsilon_0}$

If the answer is negative then force is repulsive,

Electric Field

We can take the idea further and imagine that we can measure the strength of the force between the charge at any point in space. Imagine we have we have a positive charge that has magnitude of 1. If we put this in the force equation we see that the force depends only on the magnitude of the charge we are testing. We call this the electric field strength.

Furthermore we can plot the resulting field either by joining the points that have the same strength or we can choose a color for each value.

When we add charges we can see there are some interesting field patterns.

Electrical Potential

Analogy with gravitational potential energy.

We are familiar with the idea of gravitational potential energy. When we raise a mass above the ground it has a certain potential energy which depends on the mass x height x gravitational acceleration.

Electrical charges produce an electrical potential. If you push two charges together you will have to do work to bring them closer together. If you were to let them go, the potential energy will be convert to kinetic energy as the charges fly apart.

Since the potential energy depends on the difference in potential, the zero of electrical potential can be chosen at any convenient point.

$E = -\nabla \phi$

Electrical potential at infinity is often considered to be zero.

Change in electrical potential. $\Delta V = - W/q.$ or $V = -q E d/ q = E d$

In a uniform electrical field,$\Delta V = E d.$

Electric fields occur between the plates of a capacitor.

Isopotentials and Electric Field

The electric field and the potential field lines are always perpendicular to each other. Mathematically it is the gradient of the potential.

$E = - grad V = \frac{dV_x}{dx} i + \frac{dV_y}{dy} j + \frac{dV_z}{dz}$

With the potential and electrical field overlaid we can see that they are perpendicular where there cross.

The lines of potential form a conservative field. This means that the integral of the

Laplace Equation

Using the diferrential form of Gauss's law we can relate the potential to the charge density.

$\nabla^2 \phi = \frac{\rho}{\epsilon_0}$

Two Dimensional Potential

One of the nices things about complex numbers is we can define the electric field in the complex plan and with some deft manipulation using the Cauchy-Rieman equations we can get the functions for the potential and the electric field.

The two dimensional electric field is defined by

$E = E_x + i E_y$

The potential is then defined as the derivative of E

$\frac{\partial\phi}{\partial x} - i \frac{\partial\phi}{\partial y} = 0$

Laplace's equation, from the complex electrostatic potential.

$\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y} = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0$

Every holomorphic function can be separated into its real and imaginary parts, and each of these is a solution of Laplace's equation on R2

Laplaces equation is satisfied and is harmonic on any closed curve not containing a charge. It follows that there must exist a harmonic function &Psi(x,y) conjugate to &phi(x,y) such that

$\Omega(z) = \phi(x,y) + i \Psi(x,y) = 0$

$\frac{d\Omega(z)}{dz} = \frac{\partial\phi(x,y)}{\partial x} + i\frac{\partial\Psi(x,y)}{\partial y} = 0$

Using the Cauchy Rieman equations $\frac{\partial \Psi}{\partial y} = - i\frac{\partial \phi}{\partial x}$

$\frac{d\Omega(z)}{dx} = \frac{\partial\phi(x,y)}{\partial x} - i\frac{\partial \phi(x,y)}{\partial x}$

The magnitude of the electric field is given by

$|E| = |\Omega'^*| = |\Omega'|$

with the equipotentials and flux lines given by

$\phi(x,y) = \alpha$ and $\Psi(x,y) = \beta$