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In this Physics tutorial, you will learn:

- What is electric potential energy?
- What are the similarities between electric and gravitational potential energy?
- How can we calculate the electric potential energy inside a uniform electric field?
- How can we calculate the electric potential energy produced by point charges?
- How to calculate the total electric potential energy of a system of two or more charges?

In Mechanics, we have discussed about the concept of potential energy as a kind of stored energy that can be used to do work if necessary. We can solve many problems using the law of conservation of energy, which is a fundamental law in physics. In fact, we can solve them even when we have no idea about the forces acting on the system. In the same way, we can solve many problems in electromagnetism without using the Coulomb's Law and electric field concepts. For this, we will introduce the concept of electric potential energy, which helps us explain electrical phenomena in an easier way, since energy is a scalar quantity and does not involve any direction.

In the previous tutorial, we have pointed out many similarities between gravitational and electrical nature of matter. A number of electrical concepts such as electric charge, electrostatic force, electric field etc., were explained using the analogy with the corresponding concepts discussed when dealing with gravity. We have seen that even the formulae are similar; this was very helpful in understanding the type of variation between quantities involved in electrostatics.

We can extend further this approach even for other electric-related quantities. One of them is electric potential energy. For a better understanding, we will explain this concept recalling the analogue concept of gravitational potential energy.

As you remember from Mechanics, when an object is raised at a certain height from the ground, there is some work done against gravity, which increases the object's energy. This energy is stored in the form of gravitational potential energy, which is an energy related to the distance between the Earth and the object itself. We have used the formula GPE = m × g × h to find the gravitational potential energy of objects near the Earth surface, where m is the mass of object, g is the gravitational field strength and h is the height from the ground. However, this formula is true only when h << R (R is the radius of Earth), as it represents a simplified version (a special case) of the standard formula

GPE = G × *M × m**/**d*

where G is the gravitational constant, M is the mass of Earth, m is the mass of object and d is the distance from centre of Earth to the object's position. In fact, d = R + h, but since h << R, in most cases we can approximate as d ≈ R.

When an object is raised from the ground level at a height h, it gains gravitational potential energy, as the external force used for this purpose does work against gravity. We have

W = F_{g} × ∆h

= m × g × (h - 0)

= m × g × h

= ∆GPE

= m × g × (h - 0)

= m × g × h

= ∆GPE

On the other hand, when an object falls from a height h on the ground, the gravity does work against any external force that tries to prevent the object from falling down. We have

W = F_{g} × ∆h

= m × g × (0 - h)

= -m × g × h

= -∆GPE

= m × g × (0 - h)

= -m × g × h

= -∆GPE

Thus, we can conclude that the change in gravitational potential energy of an object is equal to the negative of the work done by the gravitational force. This means when an object moves in the direction of gravitational field, it loses potential energy.

Likewise, when a positive point charge +Q is placed inside a uniform electric field produced by two oppositely charged parallel plates of distance d from each other as shown in the figure below,

the charge possesses its maximum value of potential energy when it is forced to stay at position 1, i.e. near the positive plate. This is because the positive charge cannot stay naturally at that position as it is repelled by the left (positive) plate and attracted by the right (negative) plate. Hence, when the charge is released, it moves towards the negative plate. During this process, it loses potential energy and gains kinetic energy. The electric force does positive work W on the charge because the electric force F acts in the direction of electric field E, i.e.

F*⃗* = Q × E*⃗*

This means the work done by the electric force to send the positive charge Q from position 1 to position 2, contributres in the decrease of the potential energy of the charge. Thus, we have

W = F × ∆d

= F × (d_{1} - 0)

= F × d_{1}

= -∆EPE

= -(EPE_{1} - EPE_{2} )

= EPE_{2} - EPE_{1}

= F × (d

= F × d

= -∆EPE

= -(EPE

= EPE

where EPE stands for the electric potential energy.

Given that F = Q × E, we have

W = EPE_{2} - EPE_{1}

F × d_{1} = EPE_{2} - EPE_{1}

Q × E × d_{1} = EPE_{2} - EPE_{1}

F × d

Q × E × d

Therefore, the general formula for the electric potential energy is

EPE = Q × E × d

where d is the distance of the charge Q from a point chosen as a reference (of zero potential), usually the negative plate.

The unit of electric potential enbergy is Joule [J], as all the other types of energy we have discussed earlier.

A proton is held stationary at the positive terminal of a system composed by two paralle plates charged with opposite signs, which produces a uniform electric field of magnitude 4 × 10^{5} ** N/C**. When released, the proton moves towards the negative plate which is 20 cm away from the positive one. Find the change in potential energy and the speed of proton by which it hits the negative plate. Take Q = e = + 1.6 × 10

Since the electric field inside the plates is much greater than gravitational field (remember, the object must have about 0.1 kg mass to experience a gravitational field of 1 N/kg), we can ignore this last and focus only on the electric field produced by the parallel plates.

Initially the proton has a potential energy of

EPE_{1} = Q × E × d

= e × E × d

= 1.6 × 10^{ - 19} C × 4 × 10^{5} *N**/**C* × 0.2 m

= 1.28 × 10^{ - 14} N × m

= 1.28 × 10^{ - 14} J

= e × E × d

= 1.6 × 10

= 1.28 × 10

= 1.28 × 10

Since we take the negative plate as a reference position, the electric potential energy of proton when it goes to the negative plate is zero, as d = 0. Therefore, the change in electric potential energy of proton is

∆EPE = EPE_{2}-EPE_{1}

= 0 -1.28 × 10^{ - 14} J

= -1.28 × 10^{ - 14} J

= 0 -1.28 × 10

= -1.28 × 10

As for the speed of proton when it hits the negative plate, we can use the energy conservation concept. Thus, since the entire initial potential energy of proton converts into kinetic energy at the moment it hits the negative plate, we can write

EPE_{1} = KE_{2}

EPE_{1} = *m*_{p} × v^{2}*/**2*

EPE

Therefore, the speed of electron when it hits the negative plate is

v = √*2 × EPE*_{1}*/**m*_{p}

=√*2 × 1.28 × 10*^{-14}*/**1.67 × 10*^{ - 27}

=√**1.533 × 10**^{1}3

=3.9 × 10^{6} m/s

=√

=√

=3.9 × 10

The formula EPE = Q × E × d is true only in some specific cases, when the electric field is uniform (for example between two oppositely charged parallel plates). This is similar to the formula GPE = m × g × h which is true only when the object is inside a uniform gravitational field (near the Earth surface for example). However, the formula EPE = Q × E × d cannot be used for calculating the potential energy of a charge when it is placed inside the electric field of another point charge, as in this case the field is not uniform (it is greater near the charge location as the electric field lines are closer at that part). This is similar to when objects are at a considerable height from the ground, where we must use the standard formula of gravitational potential energy

GPE = G × *M × m**/**d*

Likewise, the calculation of elastic potential energy produced by a point charge reqires a similar formula, because the field is not uniform. Thus, from the similarities between gravitation and electrostatics, we can write k (or 1/4πϵ^{0}) instead of G, Q_{1} and Q_{2} instead of M and m, and r instead of d in the formula of gravitational potential energy and obtain the corresponding formula for the electric potential energy produced by a point charge:

EPE = k × *Q*_{1} × Q_{2}*/**r*

The only difference is that EPE can be both positive and negative, depending on the sign of charges involved, unlike the GPE, which is always positive.

Electric potential energy can be defined in terms of work done by the electric forces. For example, the electric potential energy of a system composed by two like point charges is equal to the work done by the electric force to move one of the charges from the distance r to infinity.

On the other hand, when the charges are of opposite sign, the electric potential energy is equal to the negative value of the work done by the electric force to move the charge - Q from the distance r to infinity. This is because at infinity the potential energy is zero, as the value of any number divided by infinity is zero. The graph below shows the electric potential energy as a function of distance for both like and unlike charges.

Obviously, in the figure there are the graphs of two different functions: one for two like charges and the other for two unlike charges. Both graphs are hyperbolae, as the variation between PE and r is of inverse type.

Two point charges Q_{1} = 30 μC and Q_{2} = -20 μC are placed 2 m apart. How does the potential energy change if we increase the distance between charges to 6 m?

We can use the standard formula of electric potential energy

EPE = k × *Q*_{1} × Q_{2}*/**r*

to calculate the potential energy between the two charges involved. Thus, giving that 30 μC = 3 × 10^{-5}C and -20 μC = -2 × 10^{-5}C we obtain after substitutions,

EPE = 9 × 10^{9} × *3 × 10*^{-5} × (-2) × 10^{-5}*/**2*

=-2.7 J

=-2.7 J

When the distance increases to 6 m, we obtain for the new electric potential energy:

EPE=k × *Q*_{1} × Q_{2}*/**r*

= 9 × 10^{9} × *3 × 10*^{-5} × (-2) × 10^{-5}*/**6*

= -0.9 J

= 9 × 10

= -0.9 J

This means electric potential energy has increased as EPE of unlike charges increases when we separate them further.

We can use the same approach even when there are more than two charges, especially when they are not collinear. Let's consider an example in this regard.

Three charges are placed at the three vertices of an equilateral triangle as shown in the figure.

- What is the electric potential energy of the charge 3Q at the point B?
- What is the total electric potential energy of the system?

a) The potential energy of any charge at a given point is the algebraic sum of all individual potential energies due to each charge. No direction is involved, as potential energy is scalar. Thus,

EPE_{B} = EPE_{A-B} + EPE_{C-B}

EPE_{B} = k × *Q*_{a} × Q_{B}*/**r*_{a}B + k × *Q*_{C} × Q_{B}*/**r*_{CB}

= k ×*Q*_{a} × Q_{B} + Q_{C} × Q_{B}*/**r*

= k ×*-2Q) × 3Q + Q × 3Q**/**r*

= k ×*-3Q*^{2}*/**r*

EPE

= k ×

= k ×

= k ×

b) The electric potential energy of system is the algebraic sum of each pair of potential energies. Thus,

EPE_{tot} = EPE_{A-B} + EPE_{C-B} + EPE_{A-C}

= k ×*Q*_{a} × Q_{B}*/**r*_{AB} + k × *Q*_{C} × Q_{B}*/**r*_{CB} + k × *Q*_{a} × Q_{C}*/**r*_{AC}

= k ×*Q*_{a} × Q_{B} + Q_{C} × Q_{B} + Q_{a} × Q_{C}*/**r*

= k ×*(-2Q) × 3Q + Q × 3Q + (-2Q) × Q**/**r*

= k ×*-5Q*^{2}*/**r*

= k ×

= k ×

= k ×

= k ×

When a positive point charge +Q is placed inside a uniform electric field produced by two oppositely charged parallel plates of distance d from each other, the charge possesses its maximum value of potential energy when it is forced to stay near the positive plate. This is because the positive charge cannot stay naturally at that position as it is repelled by the left (positive) plate and attracted by the right (negative) plate. Hence, when the charge is released, it moves towards the negative plate. During this process, it loses potential energy and gains kinetic energy. The electric force does positive work W on the charge because the electric force F acts in the direction of electric field E, i.e.

F*⃗* = Q × E*⃗*

This means the work done by the electric force to send the positive charge Q from the positive to the negative plate, contributres in the decrease of the potential energy of the charge. Thus, we have

W = F × ∆d

= -∆EPE

= EPE_{2} - EPE_{1}

= -∆EPE

= EPE

where EPE stands for the electric potential energy.

Given that F = Q × E, we have

W = EPE_{2} - EPE_{1}

F × d_{1} = EPE_{2} - EPE_{1}

Q × E × d_{1} = EPE_{2} - EPE_{1}

F × d

Q × E × d

Therefore, the general formula for the electric potential energy is

EPE = Q × E × d

where d is the distance of the charge Q from a point chosen as a reference (of zero potential), usually the negative plate.

The unit of electric potential enbergy is Joule [J], as all the other types of energy.

The formula EPE = Q × E × d is true only in some specific cases, when the electric field is uniform (for example between two oppositely charged parallel plates). For point charges, we use another formula to calculate the electric potential energy:

EPE = k × *Q*_{1} × Q_{2}*/**r*

Electric potential energy can be defined in terms of work done by the electric forces. For example, the electric potential energy of a system composed by two like point charges is equal to the work done by the electric force to move one of the charges from the distance r to infinity.

On the other hand, when the charges are of opposite sign, the electric potential energy is equal to the negative value of the work done by the electric force to move the charge - Q from the distance r to infinity. This is because at infinity the potential energy is zero, as the value of any number divided by infinity is zero.

We can use the same approach even when there are more than two charges, especially when they are not collinear.

The potential energy of any charge at a given point is the algebraic sum of all individual potential energies due to each charge. No direction is involved, as potential energy is scalar.

1) What is the speed by which an electron initially at rest at position 1 as shown in the figure hits the plate at the position 2? Take the mass of electron equal to 9.1 × 10^{-31} kg and the charge of electron equal to - 1.6 × 10^{-19} C.

- 2.81 × 10
^{7}m/s - 2.56 m/s
- 2.56 × 10
^{7}m/s - 1.68 × 10
^{7}m/s

**Correct Answer: D**

2) Two charges Q_{1} = 40 μC and Q_{2} = 20 μC are placed at a distance r from each other. What is this distance if the electric potential energy of the system of charges is 3 J?

- 2.4 m
- 2.4 cm
- 27 cm
- 2.7 m

**Correct Answer: A**

3) What is the electric potential energy of the system of charges shown in the figure in terms of Q and r?

- 12kQ
_{2}/r - 14kQ
_{2}/r - 17kQ
_{2}/rV - 11kQ
_{2}/r

**Correct Answer: C**

We hope you found this Physics tutorial "Electric Potential Energy" 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 Potential.

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