# What is the main point in defining electric potential?

What I mean is that as the electric field is a necessity for us to define it because it has practical applications and similarly we need electric potential energy because it is the energy we require to move a unit charge from one place to another. So, when we have everything, why do we need to define electric potential? I am not understanding its practical use.

• Welcome to physics SE. Good question, should be probably answered here, I didnt find. – jaromrax Jan 18 at 15:24
• 75% of electrical engineering all day, every day, is analyzing circuits in terms of potential and currents, and never looking at fields. So at least for us EE's, it's immensely practically useful to work with potentials. (Obviously antenna engineers are the major exception to this, and even they might rather look at vector potentials than fields directly) – The Photon Jan 18 at 18:54

The electric potential is a scalar function, $$\Phi(x,y,z)$$, and the electric field is a vector field (three scalar functions), $$\mathbf{E}(x,y,z) = E_x(x,y,z)\hat{i}+E_y(x,y,z)\hat{j}+E_z(x,y,z)\hat{k}$$. So working in the scalar potential is easier, and we can always derive the electric field from it, by $$\mathbf{E} = -\nabla \Phi$$ (for the electrostatic case).

• In addition, one can get the potential energy of a system simply by specifying where the other charges are. From a practical point of view it encodes both the field and the (potential) potential energy. – garyp Jan 18 at 14:39

To elaborate on Mr Galalgher's answer,

In electro statics, two of maxwell's equations will suffice:

$$\nabla\cdot\vec{E}=\rho/\epsilon_0$$, equivalent to Coulomb's law, and, since we are only worried about static charge distributions there is no time varying Magnetic field:

$$\nabla \times \vec{E}=0$$

By a vector identity theorem, the second equation allows us to assert that $$\vec{E}=-\nabla V$$ for some scalar V.

Plug this into the first equation, we get:

$$\nabla^2V=-\rho/\epsilon_0$$.

This is a famous equation: Poisson's equation

There are several well known solutions depending on a system's symmetry.

Once you solve the differential equation for $$V$$, you can take its negative gradient to get the electric field.

Solution methods are available even if there is insufficient geometry to permit use of a Gaussian Surface, making it a more general procedure. There are also associated linear algebra techniques that make numerical solutions faster than a straight forward integration.

While you might not see electric potential per se used that often we use the difference in electric potential all the time. The potential difference is commonly called voltage. If you need to decide between a 6 volt battery or a 12 volt battery you are making a choice based on each the difference in electric potential between the terminals for each.

In quantum mechanics and classical mechanics the potential is the relevant quantity. Also on electronics the potential is used and never the electric field. Thinking in terms of energy - and momentum - is more convenient than thinking in terms of force.