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In this Physics tutorial, you will learn:

- What is electric potential (definition, formula, unit)
- What is potential difference? How is it related to the work done by the electric force?
- What are the similarities between gravitational and electric potential and potential energy?
- What are equipotential surfaces? What is the potential difference in any two points of equipotential surfaces?
- What is the common potential of a number of charged objects connected with each other through conducting wires?
- How to calculate the electric potential of a charged sphere? How does it change in different positions?
- How to deal with motion caused on a charge when it is inserted inside a uniform electric field?

In the previous tutorial, we have explained the concept of electric potential energy. Does it depend on the amount of test charge Q_{0} placed inside the electric field produced by the charge Q?

Is it possible to identify any quantity that does not depend on the amount of test charge, but which can represent quantitatively the electric field properties?

In this tutorial, we will focus precisely on this i.e. in dealing with a specific quantity representing the electric field, which is independent on the external factors such as the test charge. This quantity is known as electric potential.

In all quantities discussed so far in this section, a test charge denoted by Q_{0} was included in the formula, as it was necessary to determine and write the formulae. For example, the electric force F exerted by a charge Q on a test charge Q_{0} placed at distance r from it, is

F = k × *Q × Q*_{0}*/**r*^{2}

Also, the electric field E produced by the charge Q at a distance r from a test charge Q_{0} requires the presence of this test charge, i.e.

E = *F**/**Q*_{0}

= k*Q**/**r*^{2}

= k

despite the charge Q_{0} does not appear in the final version of the electric field formula.

Finally, the electric potential energy EPE involves the presence of the test charge Q_{0} as

EPE = k × *Q × Q*_{0}*/**r*

What if we look for a new quantity that characterizes let's say the electric field E or the electric potential energy EPE produced by the charge Q at a distance r from it but which is independent from the test charge Q_{0}? It would be very helpful as no external factors such as the test charge Q_{0} were needed anymore. This quantity would be an intrinsic property of the original charge alone.

Such a quantity exists and it is known as the **electric potential**, V. It is defined as **the electric potential energy per unit charge**.

For example, electric potential at the point A in the figure is V_{A}.

Although the potential at the point A is independent from the test charge, we can place a test charge at this point to calculate the magnitude of electric potential.

Therefore, from definition of electric potential and based on the above figures, we obtain the following formula:

V_{a} = *EPE*_{a}*/**Q*_{0}

The formula of electric potential in terms of charge Q and distance r therefore is

V = *EPE**/**Q*_{0}

=*k × **Q × Q*_{0}*/**r**/**Q*_{0}

=

Thus, we obtain for the electric potential V:

V = k × *Q**/**r*

As you see, electric potential V is independent from the test charge Q_{0} as this does not appear in the final formula.

The unit of electric potential is Volt, V. From the formula derived from definition of electric potential, it is easy to conclude that

1V = 1 *J**/**C*

What is the electric potential produced by a 12 μC point charge at a distance of 30 cm from it?

We have: Q = 12 μC = 1.2 × 10^{-5} C and r = 30 cm = 0.3 m. Thus,

V = k × *Q**/**r*

=*9 × 10*^{9} × 1.2 × 10^{-5}*/**0.3*

= 36 × 10^{4} V

= 360 000 V

=

= 36 × 10

= 360 000 V

First, we must point out the fact that positive charges placed inside an electric field move from places with higher potential to places with lower potential. Negative charges move inversely, i.e. from places with lower potential to those with higher potential. In this way, a potential difference ΔV is produced. This potential difference indicates that a movement of electric charges exists in that specific region.

Now let's turn again to the concept of electric potential energy discussed in the previous tutorial. As stated earlier, electric potential energy (as all the other types of potential energies) depends on the reference point, which we consider as a point with zero potential. It may be any of plates (usually the negative) in a system composed by two parallel plates charged with opposite signs, the location of a negative point charge and so on.

Like in Kinematics in which the difference in position is more important than the specific position at any instant, here we are more interested in the difference in potential energy or in potential than in the electric potential energy or electric potential at any instant.

Let's assume a positive charge +Q is placed inside a uniform electric field E as shown in the figure below.

When the charge moves from a to b, this is due to the action of an external force F, which does positive work because the charge moves in the opposite direction of electric field that causes an increase in its potential energy. Thus, we can write:

W_{ab} = ∆EPE

Obviously, the charge will have a difference in potential between the points a and b which is denoted by ΔV (please note that some Physics textbooks denote potential difference as U). Thus, we can write

U = ∆V = V_{b} - V_{a}

where V_{b} is the electric potential at the final point _{b} and V_{a} is the electric potential at the initial point a.

From the meaning of electric potential, we know that

V_{b} = *EPE*_{b}*/**q*

and

V_{a} = *EPE*_{a}*/**q*

where EPE_{b} and EPE_{a} are the electric potential energies of the charge q at points b and a respectively.

Therefore, we can write

U = ∆V

=*EPE*_{b}*/**q* - *EPE*_{a}*/**q*

=*∆EPE*_{a}b*/**q*

=

=

Thus, we obtain

∆V = *W*_{ab}*/**q*

The last equation is the mathematical expression of the definition of electric potential difference, i.e.

**Potential difference between two points is defined as the work done by external forces on a positive charge to move it from the initial to the final position.**

In other words, potential difference is the work per unit charge (or the change in potential energy per unit charge).

When rearranging the above formula to represent the work done on the charge to make it move from a to b, we obtain

W_{a}b = q × (V_{b} - V_{a})

= q × ∆V

= q × U

= q × ∆V

= q × U

A 5 μC charge is displaced from the point a of potential 500 V to the point b of potential 100 V.

- What is the work done by the external force during this process?
- What is the work done by the electric force during this process?
- The magnitude of electric field at the points A, B, C and D.
- The potential differences ΔV
_{CD}, ΔV_{BD}and ΔV_{AD}. - 0.8
*N**/**C* - 80
*N**/**C* - 7.2
*N**/**C* - 720
*N**/**C* - 9kQ/5R
- 8kQ/5R
- 3kQ/5R
- kQ/2R
- 0.77 s
- 0.60 s
- 0.35 s
- 0.12 s
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a) Since the electric potential decreases, the potential difference is negative as

∆V = V_{b} - V_{a}

= 100 V - 500 V

= - 400V

= -4 × 10^{2} V

= 100 V - 500 V

= - 400V

= -4 × 10

Given that the charge is positive (Q = 5 μC = 5 × 10^{-6} C), we have for the work by the external forces:

W_{ext} = Q × ∆V

= 5 × 10^{-6} C × (-4) × 10^{2} V

= -2 × 10^{-3} J

= 5 × 10

= -2 × 10

b) The negative result shows the work is done by the electric force, not by the external forces. Hence, we obtain the value 2 × 10^{-3} J for the work done by the electric force.

As stated above, when a positive charge moves in the direction of electric field, the electric forces do positive work. As a result, the electric potential energy of the positive charge decreases. This is similar to when a mass is falling from a height h. In this case, the gravity does positive work and therefore, the gravitational potential energy decreases.

On the other hand, when a force is exerted in the opposite direction of field lines, some work must be done by an external force against the electric field, which brings an increase in the electric potential energy of the charge, just like when we lift an object at a height h, where we increase the GPE of object by doing work against gravity on it.

Electric potential is similar to gravitational potential φ, which is a concept that is not very common in daily life. Using the analogy between gravitation and electricity, we can find the formula of gravitational potential at a distance r from a body of mass M:

φ = G *M**/**r*

where G is the gravitational constant.

Let's consider two charged conducting spheres of different potentials V_{1} and V_{2} as shown in the figure.

If we connect the two spheres by a conducting wire, an electric field is generated inside the wire as the spheres are at different potentials. Electric charges start flowing from the sphere with the highest potential to that with the lowest potential. This is similar to water when it flows from a higher to a lower level. In this way, the work done by the electric forces causes a charge transfer. This process continues until both spheres reach the same value of potential, i.e. until the potential difference becomes zero. Then, the charges flow stops.

If we take the radii of the above spheres as r_{1} and r_{2}, it is obvious that they share the total charge in proportion to their radii. After the contact, the spheres will have the charges q_{1} and q_{2}. From the law of charge conservation, we can write:

Q_{1} + Q_{2} = q_{1} + q_{2}

When electric potentials of the two spheres become equal, we have

V_{1} = V_{2}

or

k × *q*_{1}*/**r*_{1} = k × /*q*_{2}*/**r*_{2}

Thus, we can write

q_{2} = *r*_{2}*/**r*_{1} × q_{1}

Therefore, the law of charge distribution is written as

Q_{1} + Q_{2} = q_{1} + *r*_{2}*/**r*_{1} × q_{1}

or

Q_{1} + Q_{2} = q_{1} × (1+*r*_{2}*/**r*_{1} )

Rearranging for q_{1}, we obtain

q_{1} = *Q*_{1}+Q_{2}*/**1 + **r*_{2}*/**r*_{1}

or

q_{1} = *Q*_{1} + Q_{2}*/**r*_{1} + r_{2} × r_{1}

Similarly, for q_{2} we obtain

q_{2} = *Q*_{1} + Q_{2}*/**r*_{1} + r_{2} × r_{2}

Thus, the common potential of each sphere after the contact, is

V_{com} = k × *q*_{1}*/**r*_{1}

= k ×*Q*_{1} + Q_{2}*/**r*_{1} + r_{2}

= k ×

The last formula can be generalized for more than two charges as well, i.e.

V_{com} = k × *Q*_{1} + Q_{2} + ... + Q_{n}*/**r*_{1} + r_{2} + ... + r_{n}

Calculate the work done when a charge of Q_{1} = 4 μC is moved from the point a to the point b under the effect of a Q_{2} = 80 μC charge as shown in the figure.

The work done by electric forces is positive. This means the potential energy of the system decreases as both charges are positive. We have r_{1} = 0.3 and r_{2} = 0.4 m. Thus,

W = Q_{1} × ∆V

= Q_{1} × (V_{b} - V_{a} )

= Q_{1} × k × *Q*_{2}*/**r*_{2} - k × *Q*_{2}*/**r*_{1}

= k × Q_{1} × Q_{2} × *1**/**r*_{2} - *1**/**r*_{1}

= 9 × 10^{9} × 4 × 10^{-6} × 80 × 10^{-6} × *1**/**0.4* - *1**/**0.3*

= 36 × 80 ×*-10**/**12* × 10^{-3} J

= -2.4 J

= Q

= Q

= k × Q

= 9 × 10

= 36 × 80 ×

= -2.4 J

In the last two tutorials, we assumed the electric field between two parallel plates charged at opposite sign as uniform, especially when the distance d between the plates is small. As stated earlier, when we place a positive charge Q to the positive plate, its potential energy is at maximum. Thus,

EPE = Q × E × d

For convenience, we can assume the potential at the negative plate as zero. As a result, the charge's potential energy at the negative plate is zero as well.

We know that the potential difference is

∆V = */**Q*

=*EPE - 0**/**Q*

=*Q × E × d**/**Q*

= E × d

=

=

= E × d

If expressed in terms of E, the last formula becomes

E = *∆V**/**d*

From the above formula, we obtain the SI unit of electric field, which is **Volt/metre** [** V/m**], as the potential difference is measured in volts and distance in metres. Remember, in the previous tutorial we used another unit form electric field: Newton per Coulomb [

1 *V**/**m* = 1 *N**/**C*

The reason why the standard unit of electric field is ** V/m** and not

The formula E = ΔV / d expresses how large the potential difference in the unit of charge must be in order to produce a given electric field. Thus, if in a given region the electric field is zero, the potential in that region does not change (is constant).

Electric potential decreases when we move in the direction of electric field. Thus, for a point charge as the one shown in the figure, we have V_{1} > V_{2} > V_{3}.

As seen in the previous paragraphs, whatever type of electric field we may consider, whether uniform or non-uniform, there exist regions in these fields that have the same potential. For example, all points that have the same distance from any of plates in an electric field produced by two parallel plated charged oppositely, have the same potential, as shown in the figure.

This is because the movement of charges occur only in the direction of field lines where the potential changes continuously. Since no work is done to move the charge perpendicular to the field lines, the change in potential difference in this direction is zero. As a result, the potential V in all points aligned vertically in the above figure is constant.

The same thing occurs when the electric field is produced by a point or a spherical charge. Thus, all points that have the same distance from the point charge or the centre of sphere, have the same potential (are equipotential). In this way, an infinite number of concentric spheres whose surfaces are equipotential does result.

Two parallel plates are connected to a battery of 12 V potential difference, as shown in the figure.

If the distance between plates is 60 cm, find:

a) Electric field at any point between two parallel plates is constant. It is produced by the battery whose potential difference is ΔV = 12 V. Since the distance between the plates is d = 60 cm = 0.60 m, we have for the electric field at points A, B, C and D:

E = *∆V**/**d*

=*12 V**/**0.60 m*

= 20*V**/**m*

=

= 20

b) The points C and D are located at the same equipotential line. Thus, VC = VD. This means ΔV_{CD} = 0.

Based on the above reasoning, we have ΔV_{BD} = ΔV_{BC}. Given that d_{BD} = 20 cm = 0.20 m and d_{AD} = 40 cm = 0.40 m, we obtain

∆V_{BD}=E × d_{BD}

=20*V**/**m* × 0.20 m

=4 V

=20

=4 V

and

∆V_{AD} = E × d_{AD}

= 20*V**/**m* × 0.40 m

= 8 V

= 20

= 8 V

As explained in the previous tutorials, the electric field inside a charged sphere is zero because all charges are distributed uniformly on the surface only. This is because a sphere can naturally have only one type of extra charge, and they repel each other as far as possible as explained earlier.

Since E = ΔV / d, the potential difference inside the sphere is zero for any d as well. This means the potential inside the sphere is the same everywhere; it is equal to the potential on the outer surface, i.e.

V = k × *Q**/**R*

where Q is the amount of charge the sphere carries and R is its radius.

The potential for points outside the sphere is

V = k × *Q**/**r*

where r > R is the distance from the centre of sphere to the given point. In this case, the charge contained by the sphere behaves as if it was entirely concentrated at centre of the sphere. With the increase of distance from centre, the potential decreases, as electric potential and distance vary inversely.

Below, a graph expressing the relationship between electric potential and distance from the centre for a charged sphere is shown.

A metal sphere has a charge of Q. What is the relationship between the electric potentials at point K, L, M and N shown in the figure?

The electric potential at any point inside the sphere and on its surface is

V = k × *Q**/**R*

Here, R = 2x. Therefore, the potential at points K and L is

V_{K} = V_{L} = k × *Q**/**2x*

The points M and N are equipotential as r_{M} = r_{N} = 3x = 3R/2. Hence,

V_{M} = V_{N} = k × *Q**/**3x*

Therefore, we have

V_{M} = V_{N} > V_{M} = V_{N}

Or

V_{M} = V_{N} = *3**/**2* V_{M}

and

V_{M} = V_{N} = *3**/**2* V_{N}

An electric charge Q experiences an electric force F_{e} when it is inserted inside an electric field. As a result, it experiences an acceleration a, which based on the Newton's Second Law is

a = *F*_{e}*/**m*

where m is the mass of the charged particle.

If the charge is too small, the effect of gravity is not taken into account, so the charge will move linearly, in the direction of electric field as shown in the figure.

Given that F_{e} = Q × E, we can write for the acceleration caused by the electric force

a = *Q × E**/**m*

A proton starts moving from the positive towards the negative plate of a system composed by two parallel metal plates charged oppositely as those shown in the figure above. What is the velocity by which the proton collides with the negative plate if the distance between the plates is 20 cm? Take the mass of proton mp = 1.67 × 10^{-27} kg and the charge of electron Q = 1.6 × 10^{-19} C.

First, we must calculate the electric field E between the plates. We have

E = *∆V**/**d*

=*24 V**/**0.2 m*

= 120*V**/**m*

= 120*N**/**C*

=

= 120

= 120

Now, let's calculate the acceleration caused by the electric force on the proton. We have

a = *Q × E**/**m*

=*1.6 × 10*^{-19} C × 120 *N**/**C**/**1.67 × 10*^{-27} kg

= 1.15 × 10^{1}0 m/s^{2}

=

= 1.15 × 10

Therefore, using the kinematic formula

v^{2} - v^{2}_{0} = 2 × a × d

we find after substitutions for the speed v by which the proton hits the negative plate:

v = √**2 × a × d**

= √**2 × 1.15 × 10**^{10} × 0.2

= √**0.46 × 10**^{10}

= 6.78 × 10^{4} m/s

= √

= √

= 6.78 × 10

In some cases, especially when the charged object is quite heavy, the gravitational force cannot be neglected. As a result, there is a combination of two forces, electric and gravitational, which determine the trajectory of the charge moving inside a uniform field. Let's consider an example in this regard.

A 2 g metal sphere carrying a charge of 40 μC is inserted inside a uniform electric field as shown in the figure.

What is the potential difference between the plates if the charge remains stationary in the position shown? Take g = 9.81 N/kg.

Since there is equilibrium, we have

F_{g} = F_{e}

m × g = Q × E

m × g = Q ×*∆V**/**d*

m × g = Q × E

m × g = Q ×

Thus,

∆V = *m × g × d**/**Q*

=*2 × 10*^{-3} kg × 9.81 *N**/**kg* × 0.3 m*/**4 × 10*^{-5} C

= 147.15 V

=

= 147.15 V

Electric potential is defined as **the electric potential energy per unit charge**. It is independent from the test charge.

The formula of electric potential in terms of charge Q and distance r is

V = k × *Q**/**r*

The unit of electric potential is **Volt**, V. From the formula derived from definition of electric potential, it is easy to conclude that

1V = 1 *J**/**C*

Positive charges placed inside an electric field move from places with higher potential to places with lower potential while negative charges move in the opposite direction. As a result, a potential difference ΔV is produced. It indicates that a movement of electric charges does exist in that specific region.

**Potential difference between two points is defined as the work done by external forces on a positive charge to move it from the initial to the final position.**

In other words, potential difference is the work per unit charge (or the change in potential energy per unit charge).

∆V = *W*_{ab}*/**q*

When a positive charge moves in the direction of electric field, the electric forces do positive work. As a result, the electric potential energy of the positive charge decreases. This is similar to when a mass is falling from a height h. In this case, the gravity does positive work and therefore, the gravitational potential energy decreases. On the other hand, when a force is exerted in the opposite direction of field lines, some work must be done by an external force against the electric field, which brings an increase in the electric potential energy of the charge, This is similar to when we lift an object upwards at a height h, in which we increase the gravitational potential energy of the object by doing work against gravity on it.

Whatever type of electric field we may consider - whether uniform or non-uniform - there exist some regions in these fields that have the same potential. For example, all points that have the same distance from any of plates in an electric field produced by two parallel plated charged oppositely, have the same potential,

The same thing occurs when the electric field is produced by a point or a spherical charge as well. All points that have the same distance from the point charge or the centre of sphere, have the same potential (are equipotential). In this way, an infinite number of concentric spheres whose surfaces are equipotential does result.

Electric potential produced by a number n of charged spheres when they are connected through conducting wires is

V_{com} = k × *Q*_{1} + Q_{2} + ... + Q_{n}*/**r*_{1} + r_{2} + ... + r_{n}

For an individual charged sphere of radius R, the potential inside the sphere is

V = k × *Q**/*R

The potential for points outside the sphere is

V = k × *Q**/**r*

where r > R is the distance from the centre of sphere to the given point.

A charge Q experiences an electric force F_{e} when it is inserted inside an electric field. As a result, it experiences an acceleration a, which based on the Newton's Second Law is

a = *F*_{e}*/**m*

where m is the mass of the charged particle.

When the charged object is quite heavy, the gravitational force cannot be neglected. As a result, there is a combination of two forces, electric and gravitational, which determine the trajectory of the charge moving inside a uniform field.

1) What is the electric field between the plates for the system shown in the figure?

**Correct Answer: B**

2) What is the potential difference between the points A and B in figure in terms of electric charge Q and radius of sphere R?

**Correct Answer: C**

3) A 200 μC charge of mass 0.5 g is initially at rest on the positive plate of the system shown in the figure.

How many seconds does it take to the charge to reach the negative plate? (The answer is rounded to two decimal places).

**Correct Answer: A**

We hope you found this Physics tutorial "Electric Potential" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Electrostatics with our Physics tutorial on Electric Flux. Gauss Law.

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