Physics Lesson 15.3.1 - Recap on Electric Potential and Potential Difference of a Charged Object

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Welcome to our Physics lesson on Recap on Electric Potential and Potential Difference of a Charged Object, this is the first lesson of our suite of physics lessons covering the topic of Electric Potential Difference (Voltage). Ohm's Law, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Recap on Electric Potential and Potential Difference of a Charged Object

In the previous chapter (section), we have explained the concept of electric potential produced by a charged object. Electric potential was defined as the electric potential energy per unit charge. The equation of electric potential was

where Q is the charge that produces the electric field and r is the distance from that charge, at which we want to calculate the electric potential V.

We also gave the unit of electric potential, i.e. Volt [V]. Given that the electric potential of a trial charge Q₀ located inside the electric field produced by the charge Q at the distance r from it, is

where W is the electric potential energy of the trial charge at the given distance, we obtain for the unit of electric potential

However, the definition of the electric potential given earlier (electric potential energy per unit charge) looks more a mathematical than a physical definition. It does not gives so much info about the importance of electric potential as a quantity. Therefore, in many textbooks, a more practical definition for the electric potential is used. Thus,

"Electric potential V, is the work done by an electric force to bring a positive test charge Q₀ from infinity at a given distance r from another positive charge Q."

Now, the purpose of the electric potential existence is much clearer. Electric field produced by a charged object is different at different distances from the object, but it is constant at all points that are at the same distance from the charged object. This property leads to the concept of equipotential surfaces, as shown in the figure below.

From the figure, it is obvious that VB = VC as both these points have the same distance from the charge that generates the field. We say two test charges located at the points B and C are on an equipotential surface.

On the other hand, the point A is on another equipotential surface, which has a different value from that of the points B and C as the distance from the central charge to each of these points is different. As a result, the positive charge at middle of the sketch, which produces the field E, is able to move a positive test charge from A to B (or a negative test charge from B to A). However, it cannot move a charge from B to C or from C to B, as the potential in these points is equal and the work in the direction of the field lines is zero.

Thus, it is clear that a charged object can do work on a test charge only if it changes the value of electric potential of the test charge. In this way, a potential difference ΔV = V₂ - V₁ is produced, where V₁ and V₂ are the values of initial and final electric potential of the test charge. If the corresponding distances from the centre of the field are R₁ and R₂ respectively, we obtain for the potential difference in the two given points:

Here, we must consider all possible cases.

1. If a positive test charge comes from the infinity (R₁ = ∞) to a distance R₂ from the point charge which produces the field, it (the test charge) experiences a potential difference
   ∆V = k ∙ Q ∙ (1/R₂ - 1/∞)
   = k ∙ Q/r₂
2. The work done from the central charge in this case is negative as it acts against the field lines. Thus, we have
   W = -Q ∙ ∆V
   = -Q ∙ V₂
3. If a negative test charge comes from the infinity (R₁ = ∞) to a distance R₂ from the point charge which produces the field, it (the test charge) experiences a potential difference
   ∆V = -k ∙ Q ∙ (1/R₂ -1/∞)
   = -k ∙ Q/r₂
4. The work done from the central charge in this case is positive as it acts against the field lines. Thus, we have
   W = Q ∙ ∆V
   = Q ∙ V₂
   (the sign minus in the potential difference comes due to the interaction of two opposite charges).
5. If a positive charge is brought from a distance R₁ to a distance R₂ from the point charge that produces the field, the potential difference experienced by the testy charge is
   ∆V = k ∙ Q ∙ (1/R₂ -1/R₁ )
6. If a positive test charge is brought from a distance R₁ to a distance R₂ from the point charge which produces the field, the potential difference experienced by the test charge is
   ∆V = -k ∙ Q ∙ (1/R₂ - 1/R₁ )
   = k ∙ Q ∙ (1/R₁ - 1/R₂ )

Example 1

A positive test charge is brought from R₁ = 3 m to R₂ = 1 m away from a 6 μC point charge that produces an electric field around it. What is the potential difference between the two given points? Take k = 9 × 109 N ∙ m2 / C2.

Solution 1

Using the equation of the potential difference between two points, we obtain

You have reached the end of Physics lesson 15.3.1 Recap on Electric Potential and Potential Difference of a Charged Object. There are 4 lessons in this physics tutorial covering Electric Potential Difference (Voltage). Ohm's Law, you can access all the lessons from this tutorial below.

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