10.7 Conservation of Electric Energy

When a charged object moves between two points with different electric potentials, its electric potential energy changes by $\Delta U_E = q \Delta V$. Because energy is conserved, that change in potential energy shows up as a change in kinetic energy, which is how charged particles speed up or slow down in electric fields.

Conservation of energy, the same tool you used in mechanics, now links electric potential to a charge's motion, and Unit 10 is strong practice for the free-response question that asks you to translate between representations. A scenario like a charge released near a charged object is a good test: connect the verbal description, a diagram, and the equation for the same situation.

Charges Gain Kinetic Energy as Potential Energy Drops

• The change in electric potential energy for a charge moving between two potentials is $\Delta U_E = q \Delta V$, measured in joules.
• Track signs carefully: $q$ can be positive or negative, and $\Delta V = V_{final} - V_{initial}$ can be positive or negative.
• Energy is conserved, so a drop in electric potential energy becomes an increase in kinetic energy: $\Delta K = -\Delta U_E$.
• Positive charges speed up when moving toward lower potential; negative charges speed up when moving toward higher potential.
• The electric field points toward decreasing potential, so use the field, not just terminal labels, to decide which way a charge accelerates.
• For a particle starting from rest, the kinetic energy gained equals the magnitude of $q \Delta V$, which lets you solve for final speed.

Changes in Energy Due to Electric Potential Difference

Electric Potential Energy Changes

When a charged object moves through an electric field, its electric potential energy changes. This is similar to how an object's gravitational potential energy changes when it moves up or down in Earth's gravitational field.

• The change in electric potential energy is directly proportional to both the charge of the object and the potential difference between the two locations.
• This relationship is written as: $\Delta U_E = q \Delta V$
  • $\Delta U_E$ is the change in electric potential energy (in joules)
  • $q$ is the charge of the object (in coulombs)
  • $\Delta V$ is the electric potential difference (in volts)
• The sign of the energy change depends on the sign of the charge and the direction of movement.
  • Positive charges gain potential energy when moving to higher potential.
  • Negative charges gain potential energy when moving to lower potential.

For example, when an electron (negative charge) moves from a lower electric potential to a higher electric potential, its electric potential energy decreases because $\Delta U_E = q\Delta V$ and $q$ is negative. The direction the electron naturally accelerates depends on the electric field, not just on the terminal labels. A proton pushed toward another proton gains electric potential energy as it moves against the repulsive force.

Conservation of Energy Principle

Energy cannot be created or destroyed, only transformed from one form to another. This principle governs how charged particles behave in electric fields.

When a charged particle moves between points with different electric potentials:

• The change in kinetic energy equals the negative of the change in electric potential energy.
• This is written as: $\Delta K = -\Delta U_E = -q\Delta V$
• The total mechanical energy of the object-field system stays constant: $\Delta K + \Delta U_E = 0$

This shows up in many electrical situations:

• An electron accelerating in an electric field converts electric potential energy into kinetic energy.
• A proton slowing down as it approaches another proton converts kinetic energy into electric potential energy.
• In a circuit, the electric potential energy of the charge-field system can be transformed into other forms such as thermal energy, light, or sound as charges move through circuit elements.

This idea is central to understanding charged particles in accelerators, cathode ray tubes, and other devices where charges move through potential differences.

Solving Energy Conservation Problems With Charges

Problem Solving:

• Start by writing $\Delta V = V_{final} - V_{initial}$ before plugging into $\Delta U_E = q \Delta V$. Sign mistakes here cause most wrong answers.
• Use $\Delta K = -\Delta U_E$ to connect potential energy changes to speed changes.
• For a charge starting from rest, set $\frac{1}{2}mv^2 = |q \Delta V|$ and solve for $v$.

Free Response:

• When asked to translate between representations, line up three views of the same scenario: a diagram (equipotential lines or field vectors), an energy bar chart, and the equation $\Delta U_E = q \Delta V$.
• In energy bar charts, show that any decrease in electric potential energy matches an increase in kinetic energy so the bars stay consistent with conservation of energy.
• When you explain your reasoning, state the sign of the charge, the sign of $\Delta V$, and what that means for kinetic energy.

Common Trap:

• Do not assume a charge always speeds up when moving to higher potential. That is only true for negative charges. Negative charges speed up moving toward higher potential because $q$ is negative.

Practice Problem 1: Electric Potential Energy Change

Solution

Use $\Delta U_E = q\Delta V$.

Identify the values:

• Charge of proton, q = +1.6 × 10^-19 C
• Initial potential, V₁ = +200 V
• Final potential, V₂ = +50 V

The change in potential is: $\Delta V = V_2 - V_1 = 50\text{ V} - 200\text{ V} = -150\text{ V}$

Now calculate the change in electric potential energy: $\Delta U_E = q\Delta V = (1.6 \times 10^{-19}\text{ C})(-150\text{ V}) = -2.4 \times 10^{-17}\text{ J}$

Since the change in electric potential energy is negative, the proton loses electric potential energy. By conservation of energy, its kinetic energy increases by the same amount. The proton gains 2.4 × 10^-17 J of kinetic energy as it moves from the higher potential to the lower potential.

Practice Problem 2: Conservation of Energy

Solution

Use conservation of energy. The change in kinetic energy equals the negative of the change in electric potential energy.

Identify the values:

• Mass of electron, m = 9.11 × 10^-31 kg
• Charge of electron, q = -1.6 × 10^-19 C
• Initial potential, V₁ = 0 V
• Final potential, V₂ = +5000 V
• Initial velocity, v₁ = 0 (electron starts at rest)

The change in electric potential energy is: $\Delta U_E = q\Delta V = q(V_2 - V_1) = (-1.6 \times 10^{-19}\text{ C})(5000\text{ V} - 0\text{ V}) = -8.0 \times 10^{-16}\text{ J}$

Since energy is conserved, this decrease in electric potential energy results in an equal increase in kinetic energy: $\Delta K = -\Delta U_E = 8.0 \times 10^{-16}\text{ J}$

The final kinetic energy is: $K_2 = K_1 + \Delta K = 0 + 8.0 \times 10^{-16}\text{ J} = 8.0 \times 10^{-16}\text{ J}$

Using $K = \frac{1}{2}mv^2$, solve for the final velocity: $v_2 = \sqrt{\frac{2K_2}{m}} = \sqrt{\frac{2(8.0 \times 10^{-16}\text{ J})}{9.11 \times 10^{-31}\text{ kg}}} = 4.2 \times 10^7\text{ m/s}$

The electron reaches a speed of about 4.2 × 10^7 m/s after accelerating through the potential difference.

Common Misconceptions

• Misconception: Higher potential always means more energy for a charge.
  Actually: Only for positive charges. For a negative charge, $\Delta U_E = q \Delta V$ is negative when $\Delta V$ is positive, so its potential energy drops.
• Misconception: A charge always moves toward lower potential.
  Actually: Particles move based on the force from the field. Positive charges tend toward lower potential, but negative charges tend toward higher potential.
• Misconception: Kinetic energy gained equals $\Delta V$.
  Actually: Kinetic energy gained equals the magnitude of $q \Delta V$, not $\Delta V$ by itself. You must include the charge.
• Misconception: Forgetting the sign of $\Delta V$.
  Actually: Always compute $\Delta V = V_{final} - V_{initial}$. Reversing this flips the sign of your energy change.
• Misconception: Electric potential energy and electric potential are the same thing.
  Actually: Potential energy ($U_E$) depends on the charge moved; potential ($V$) is energy per unit charge and does not depend on a test charge being present.

Related AP Physics 2 Guides

• 10.1 Electric Charge and Electric Force
• 10.2 The Process of Charging
• 10.3 Electric Fields
• 10.5 Electric Potential
• 10.6 Capacitors
• 10.4 Electric Potential Energy