# Electric potential: formula and equations, calculation, examples, exercises

The **electric potential** is defined at any point where there is an electric field, as the potential energy of said field per unit charge. Point charges and point or continuous charge distributions produce an electric field and therefore have a potential associated with them.

In the International System of Units (SI), electrical potential is measured in volts (V) and is denoted as V. Mathematically, it is expressed as:

*V = U / q _{or}*

Where U is the potential energy associated with the load or distribution and q _{o} is a positive test load. Since U is a scalar, so is the potential.

From the definition, 1 volt is simply 1 Joule / Coulomb (J / C), where Joule is the SI unit for energy and Coulomb (C) is the unit for electrical charge.

Suppose a point charge q. We can verify the nature of the field this charge produces using a small positive test charge, called q _{o} , used as a probe.

The work W required to move this small charge from point *a* to point *b* is negative of the *potential energy* difference ΔU between these points:

*W _{a → b} = -ΔU = – (U _{b} – U _{a} ) *

Dividing everything by q _{or} :

*A _{= b} / q _{o} = – ΔU / q _{o} = – (U _{b} – U _{a} ) / q _{o} = – (V _{b} – V _{a} ) = -ΔV*

Here V _{b} is the potential at point b and V _{a} is point a. The potential difference V _{a} – V _{b }is the potential of a *with respect to b and* is called V _{ab} . The order of the subscripts is important, if changed, it would represent the potential of *b in relation to a* .

__Electric potential difference__

__Electric potential difference__

From the above, it follows that:

*-ΔV = W _{a → b} / q _{or}*

Therefore:

*ΔV = -W _{a → b} / q _{or}*

Now the work is calculated as the integral of the scalar product of the electric force **F** between qeq _{or} the displacement vector d **ℓ** between points a and b. As the electric field is the force per unit charge:

*E** = F / q _{or}*

The work to transport the test load from a to b is:

This equation provides the way to directly calculate the potential difference if the electric field of the charge or the distribution that produces it is known in advance.

And also note that the potential difference is a scalar quantity, unlike the electric field, which is a vector.

**Signals and values for potential difference**

From the above definition, we observe that if **E and** d **ℓ** are perpendicular, the potential difference ΔV is zero. This does not mean that the potential at such points is zero, but simply that V _{a} = V _{b} , that is, the potential is constant.

The lines and surfaces where this happens are called *equipotentials* . For example, the equipotential lines of the field of a point charge are circles concentric to the charge. And equipotential surfaces are concentric spheres.

If the potential is produced by a positive charge, whose electric field consists of radial lines that protrude from the charge, as we move away from the field, the potential becomes smaller and smaller. Since the test charge q _{o} is positive, you feel less electrostatic repulsion the farther you are from q.

On the other hand, if the charge *q* is negative, the test charge q _{or} (positive) will have less potential as it approaches *q.*

__How to calculate electrical potential?__

__How to calculate electrical potential?__

The integral given above is used to find the potential difference and therefore the potential at a given point *b* , if the reference potential at another point *a is known.*

For example, there is the case of a point charge *q* , whose vector electric field at a point located at a distance *r* from the charge is:

**E** = kq / r ^{2 }**r**

Where k is the electrostatic constant whose value in International System units is:

k = 9 x 10 ^{9} Nm ^{2} / C ^{2} .

And vector * r * is the unit vector along the line that connects

*q*to point P.

It is substituted in the definition of *ΔV* :

The choice of point *b* is at a distance *r* from the charge and, when a → ∞ the potential is 0, then V _{a} = 0 and the previous equation remains as:

V = kq / r

Choosing V _{a} = 0 when a → ∞ makes sense, because at a point very far from the charge, it’s hard to see that it exists.

**Electrical potential for discrete charge distributions**

When there are many point charges distributed in a region, the electric potential they produce at any point P in space is calculated by adding up the individual potentials produced by each one. Thus:

V = V _{1} + V _{2} + V _{3} + … VN = ∑ V _{i}

The sum ranges from i = to N and the potential of each charge is calculated using the equation given in the previous section.

**Electrical potential in continuous charge distributions**

From the potential of a point charge, the potential produced by a charged object, with a measurable size, can be found at any point P.

For this, the body is divided into many small infinitesimal charges *dq* . Each contributes to the full potential with an infinitesimal *dV* .

All these contributions are added together through an integral and the full potential is obtained:

__Examples of electrical potential__

__Examples of electrical potential__

There is electrical potential in various devices through which electrical energy can be obtained, such as batteries, car batteries, and electrical outlets. Electrical potentials are also established in nature when there are storms.

**Batteries and Batteries**

Electric cells are stored in cells and batteries through chemical reactions inside. This occurs when the circuit is closed, allowing direct current to flow and a light bulb to light, or the car’s starter motor to run.

They are of different voltages: 1.5V, 3V, 9V and 12V are the most common.

**Exit**

Appliances that run on commercial AC power are connected to a wall outlet. Depending on the location, the voltage can be 120V or 240V.

**Tension between charged clouds and earth**

This is what happens during electrical storms, due to the movement of electrical charge through the atmosphere. It can be in the order of 10 ^{8} V.

**Van Der Graff Generator**

Thanks to a rubber conveyor belt, a frictional load is produced which builds up on a conducting sphere placed on top of an insulating cylinder. This generates a potential difference that can be several million volts.

**Electrocardiogram and EEG**

In the heart, there are specialized cells that polarize and depolarize, causing possible differences. These can be measured as a function of time using an electrocardiogram.

This simple test is done by placing electrodes on the person’s chest that can measure small signals.

Because they are very low voltages, they must be conveniently amplified and then recorded on a paper tape or viewed through a computer. The doctor analyzes the pulses for abnormalities and thus detects heart problems.

The brain’s electrical activity can also be recorded with a similar procedure called an electroencephalogram.

__Exercise solved__

__Exercise solved__

A load *Q* = – 50.0 nC is located 0.30 m from point *A* and 0.50 m from point B, as shown in the following figure. Answer the following questions:

a) What is the potential in A produced by this charge?

b) And what is the potential in B?

c) If a charge q passes from A to B, what is the potential difference by which it does so?

d) According to the previous answer, does its potential increase or decrease?

e) If q = –1.0 nC, what is the change in its electrostatic potential energy as it moves from A to B?

f) How much work does the electric field produced by Q do when the test charge moves from A to B?

**Solution for**

Q is a point charge, so its electrical potential in A is calculated by:

V _{A} = kQ / r _{A} = 9 x 10 ^{9} x (-50 x 10 ^{-9} ) / 0.3 V = -1500 V

**Solution b**

similarly

V _{B} = kQ / r _{B} = 9 x 10 ^{9} x (-50 x 10 ^{-9} ) / 0.5 V = -900 V

**Solution c**

ΔV = V _{b} – V _{a} = -900 – (-1500) V = + 600 V

**Solution of**

If the charge q is positive, its potential increases, but if it is negative, its potential decreases.

**Solution and**

*ΔV = ΔU / q _{or} → ΔU = q _{or} ΔV = -1.0 x 10 ^{-9} x 600 J = -6.0 x 10 ^{-7} J.*

The negative sign at *ΔU* indicates that the potential energy at B is less than that of A.

**Solution f**

Since W = -ΔU, the field does *+6.0 x 10 ^{-7} J* of work.