Saved calculators
Physics

Electric potential energy calculator

Report a bug

Share calculator

Add our free calculator to your website

Please enter a valid URL. Only HTTPS URLs are supported.

Use as default values for the embed calculator what is currently in input fields of the calculator on the page.
Input border focus color, switchbox checked color, select item hover color etc.

Please agree to the Terms of Use.
Preview

Save calculator

Electric potential energy

Electric potential energy is a key concept in electromagnetism, describing the energy a charged particle holds due to its position within an electric field. Here we will delve into three distinct calculations of electric potential energy, each applicable under different circumstances.

Three calculations of electric potential energy

1. Charge in an electric field

When a charge is situated within an electric field, the electric potential energy (UU) can be determined using:

U=qVU = q \cdot V

Where:

  • UU is the electric potential energy,
  • qq is the charge,
  • VV is the electric potential at the location of the charge.

This formula applies when the electric potential in a specific point and the charge magnitude are known.

Example calculation

Consider a charge of 2 μC2 \ \mu C situated in a field with an electric potential of 5 V5\ V:

U=(2×106 C)5 V=1×105 JU = (2 \times 10^{-6}\ \text{C}) \cdot 5\ \text{V} = 1 \times 10^{-5}\ \text{J}

2. Moving a charge in an electric field

When moving a charge in a uniform electric field, the potential energy change is given by:

U=qEdU = q \cdot E \cdot d

Where:

  • EE is the electric field strength,
  • dd is the displacement of the charge in the field direction.

Example calculation

If a 3 μC3\ \mu C charge moves 0.1 m0.1\ m in a field of 20 V/m20\ V/m:

U=(3×106 C)(20 V/m)0.1 m=6×107 JU = (3 \times 10^{-6}\ \text{C}) \cdot (20\ \text{V/m}) \cdot 0.1\ \text{m} = 6 \times 10^{-7}\ \text{J}

3. Interaction of two point charges

When calculating the potential energy associated with the interaction between two point charges:

U=keq1q2rU = k_e \cdot \frac{q_1 \cdot q_2}{r}

Where:

  • UU is the potential energy of interaction,
  • q1q_1 and q2q_2 are the magnitudes of the charges,
  • rr is the distance between the charges,
  • kek_e is Coulomb’s constant (8.9875×109 N m2/C2)(8.9875 \times 10^9\ \text{N m}^2/\text{C}^2).

Example calculation

For two charges q1=1 μCq_1 = 1\ \mu C and q2=2 μCq_2 = 2\ \mu C separated by 0.05 m0.05\ m:

U=(8.9875×109) ×(1×106)×(2×106)0.05=0.3595 JU = \frac{(8.9875 \times 10^9)\ \times (1 \times 10^{-6}) \times (2 \times 10^{-6})}{0.05} = 0.3595\ \text{J}

Examples and applications

Let’s explore a few intriguing examples to illustrate the calculation of electric potential energy in practical scenarios.

Example 1: A proton in a parallel plate capacitor

Consider a proton, carrying a charge of 1.602×10191.602 \times 10^{-19} C, placed within a parallel plate capacitor. The capacitor has a voltage of 12V across its plates.

Using the formula:

U=qV=(1.602×1019 C)12 V=1.9224×1018 JU = q \cdot V = (1.602 \times 10^{-19}\ \text{C}) \cdot 12\ \text{V} = 1.9224 \times 10^{-18}\ \text{J}

This energy represents the work needed to move the proton across the capacitor and is critical in understanding operations like particle acceleration and in applications like cathode ray tubes and mass spectrometers.

Example 2: Electron movement in circuit

An electron, with a charge of 1.602×1019-1.602 \times 10^{-19} C, is moved through a potential difference of 4545 volts (such as in a television screen or an oscilloscope).

U=qV=(1.602×1019 C)45 V=7.209×1018 JU = q \cdot V = (-1.602 \times 10^{-19}\ \text{C}) \cdot 45\ \text{V} = -7.209 \times 10^{-18}\ \text{J}

The negative sign indicates that the direction of the electron’s movement opposes the direction of the electric field, a fundamental principle underlying the flow of current in electronics.

Example 3: Water molecule influencing ion

A water molecule, having an induced charge due to a negatively charged ion, experiences these complex interactions in biochemical contexts. Determine the potential energy if the molecule is near a charge of magnitude 2×1019 C2 \times 10^{-19}\ \text{C} and subjected to a field strength of 1000 V/m1000\ \text{V/m} across a 0.2 m0.2\ \text{m} distance.

U=qEd=(2×1019 C)(1000 V/m)0.2 m=4×1020 JU = q \cdot E \cdot d = (2 \times 10^{-19}\ \text{C}) \cdot (1000\ \text{V/m}) \cdot 0.2\ \text{m} = 4 \times 10^{-20}\ \text{J}

This calculation is crucial in studying chemical bonding and reaction energetics.

Relevance in modern technology

Electric potential energy plays a pivotal role in various modern technologies. It’s central to the design of electric circuits, enabling the operation of batteries and capacitors. In addition, it underlies the principles of electric power generation and distribution. Devices like smartphones, computers, and electric cars rely on the effective management and conversion of electric potential energy.

Frequently asked questions

How to calculate electric potential energy for a charge in a field of 10 V/m?

Given the field strength (E=10 V/mE = 10\ \text{V/m}), charge (q=5 μC=5×106 Cq = 5\ \mu\text{C} = 5 \times 10^{-6}\ \text{C}) and distance (d=2 md = 2\ \text{m}), calculate:

U=qEd=(5×106)(10)2=1×104 JU = q \cdot E \cdot d = (5 \times 10^{-6}) \cdot (10) \cdot 2 = 1 \times 10^{-4}\ \text{J}

Why is electric potential energy important in electrical systems?

It represents stored energy that can be converted into kinetic energy or work, essential for understanding circuits and electrical devices.

What is the difference between electrostatic potential energy and electric potential energy?

Electrostatic potential energy relates to interactions between multiple charged particles; electric potential energy refers to a single charge’s energy within a field.

How many joules of energy are required to move an electron through a 100 V potential difference?

Given the electron charge (1.602×1019-1.602 \times 10^{-19} C), calculate:

U=qV=(1.602×1019)(100)=1.602×1017 JU = q \cdot V = (-1.602 \times 10^{-19}) \cdot (100) = -1.602 \times 10^{-17}\ \text{J}

What role does electric potential energy play in electric generators?

It allows the conversion of mechanical energy into electric kinetic energy, powering systems across industries and homes.