Electric Potential Examples in Physics

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Electric Potential.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Physics.

Concept Recap

The electric potential energy per unit charge at a point in an electric field. Measured in volts (V).

Electric potential is like altitude on a hill — charges 'roll downhill' from high potential to low potential, just as balls roll from high ground to low ground.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Potential is a scalar — no direction, just a number at each point. The difference in potential between two points drives current.

Common stuck point: Potential is defined at a single point, but only the difference between two points (voltage) does physical work.

Sense of Study hint: When solving an electric potential problem, identify the source charge and the distance to the point of interest. Then apply V = kQ/r — note there is no squaring of r unlike the field formula. If multiple charges are present, add their potentials as scalars (no vector addition needed).

Worked Examples

Example 1

medium
What is the electric potential 0.3 \text{ m} from a point charge of 6 \times 10^{-6} \text{ C}? Use k = 9 \times 10^9 \text{ N m}^2/\text{C}^2.

Solution

  1. 1
    Use the electric potential formula for a point charge: V = k\dfrac{q}{r}, where k = 9 \times 10^9\,\text{N m}^2/\text{C}^2.
  2. 2
    Identify given values: q = 6 \times 10^{-6}\,\text{C}, r = 0.3\,\text{m}. Compute \dfrac{q}{r} = \dfrac{6 \times 10^{-6}}{0.3} = 2 \times 10^{-5}.
  3. 3
    Multiply by k: V = 9 \times 10^9 \times 2 \times 10^{-5} = 1.8 \times 10^5\,\text{V}

Answer

V = 1.8 \times 10^5 \text{ V}
Electric potential is the electric potential energy per unit charge. Unlike the electric field (a vector), potential is a scalar quantity, making it easier to work with in many calculations.

Example 2

hard
How much work is needed to move a 3 \times 10^{-6} \text{ C} charge from a point at 200 \text{ V} to a point at 500 \text{ V}?

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

medium
An electron (q = -1.6 \times 10^{-19} \text{ C}) is accelerated through a potential difference of 1000 \text{ V}. What kinetic energy does it gain?

Example 2

hard
A uniform electric field of E = 500 \text{ N/C} exists between two parallel plates separated by d = 0.02 \text{ m}. (a) What is the potential difference between the plates? (b) How much work is done moving a proton (q = 1.6 \times 10^{-19} \text{ C}) from the negative to the positive plate?
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Background Knowledge

These ideas may be useful before you work through the harder examples.

electric fieldcoulombs law