D2-4. Electric Potential, Energy and GradientsHL Only

1. Electric Potential ($V$)

The Formula & Fundamental Principles

$$V = \frac{k q}{r} \qquad \text{where} \qquad k = \frac{1}{4\pi\varepsilon_0}$$
  • Definition: The electric potential ($V$) at a specific point in space is defined as the work done per unit charge by an external agent in bringing a small, positive test charge from infinity ($\infty$) to that exact point without acceleration.
  • Nature: It is a scalar quantity. Unlike the vector electric field ($\vec{E}$), direction does not exist for potential—only magnitude and algebraic sign matter. This significantly simplifies calculations since multi-charge systems can be evaluated using simple algebraic addition rather than vector decomposition.
  • Zero Reference Point: By universal convention, the absolute electric potential is established as exactly zero at an infinite distance away ($V = 0$ at $r = \infty$). Consequently, positive charges create a positive potential "hill" or peak, while negative charges create a negative potential "well" or valley.
  • Principle of Superposition: The total electric potential at a point due to a static collection of discrete point charges is the direct algebraic sum of the potentials generated by each individual charge independently: $$V_{\text{total}} = \sum_{i} \frac{k q_i}{r_i} = k \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3} + \dots \right)$$
  • Equipotential Contours: Regions in space where the electric potential remains completely uniform are called equipotential lines (in 2D) or surfaces (in 3D). Because $\Delta V = 0$ along any path on an equipotential surface, the net work done in moving a charge along these contours is zero ($W = q\Delta V = 0$).
  • Field Intersections ($\vec{E} \perp V$): Electric field lines always run **perpendicular** to equipotential contours at every point in space. Furthermore, field vectors point in the direction of the steepest *decrease* in electric potential, mathematically expressed via the gradient relationship: $$\vec{E} = -\nabla V \implies E_{\text{local}} \approx -\frac{\Delta V}{\Delta r}$$ Where contours are tightly compressed or bundled together, the potential changes rapidly over a short distance, indicating a highly intense local electric field.
Equipotential Contours & Electric Field of a Dipole 0V (Equipotential Plane) +q -q +25V +50V +100V -100V -50V -25V Positive Potential Negative Potential Electric Field ($\vec{E}$)

2. Electric Potential Energy ($E_p$) & Work Done ($\Delta W$)

The Formulas & Extended Concepts

$$E_p = qV = \frac{k q_1 q_2}{r}$$ $$W_{\text{ext}} = \Delta E_p = q\Delta V$$
  • Electric Potential Energy ($E_p$): The total macroscopic energy required by an external agent to assemble a system of point charges from infinite separation to their current configuration. For a multi-charge system, the total energy is the algebraic sum of the potential energy of every unique interacting pair of charges ($E_{\text{total}} = \sum \frac{kq_i q_j}{r_{ij}}$).
  • Work Done ($\Delta W$): The mechanical work done by an external force to move a charge $q$ between two points with a potential difference $\Delta V$.
    • External Work vs. Field Work: Work done by an external agent moving a charge at constant velocity is $W_{\text{ext}} = +q\Delta V$. The work done by the electric field itself during this process is exactly opposite: $W_{\text{field}} = -q\Delta V$.
    • Conservation of Energy: If a charge is released from rest and moves freely under the influence of the field, it accelerates. The electric field does positive work, converting Electric Potential Energy into Kinetic Energy ($\Delta E_k = -\Delta E_p$).
  • Motion along Equipotentials: If a charge moves along or within an equipotential surface, there is no change in potential ($\Delta V = 0$). Consequently, the work done by the electric field is strictly zero. This implies the force is perfectly perpendicular to the displacement at all times.

Sign Interpretations & Stability

  • Positive $E_p$ (Repulsive Systems): The system consists of like charges. You must do positive external work to force them together against their natural repulsion. The system is inherently unstable and will fly apart, releasing kinetic energy, if constraints are removed.
  • Negative $E_p$ (Attractive Systems): The system consists of unlike charges. They naturally attract, so the field does the work pulling them together (they fall into a "potential well"). You must supply positive external work to tear them apart.

3. Electric Potential Gradient

The Formula & Geometric Meaning

$$E = -\frac{\Delta V}{\Delta r} \qquad \text{(In 1D or uniform fields)}$$ $$\vec{E} = -\nabla V \qquad \text{(Advanced calculus form)}$$
  • The Concept: The electric field strength ($E$) at any spatial coordinate is mathematically equivalent to the negative spatial gradient of the electric potential at that point. It measures how rapidly the voltage changes over distance.
  • The Negative Sign: Electric field lines inherently map the "downhill" direction of the potential landscape. They always point in the direction of the steepest decreasing electric potential.
  • Line Spacing and Field Strength: On a contour map, where equipotential lines are clustered tightly together, $\Delta r$ is small for a given $\Delta V$. This steep gradient indicates a highly intense local electric field. Where lines are spaced far apart, the electric field is weak.
  • Orthogonality: Electric field vectors must intersect equipotential surfaces at exactly $90^\circ$ (perpendicularly). If the field had a parallel component along the surface, it would do work on charges moving along the surface, violating the definition of an equipotential zone.
Uniform Field (Parallel Plates) ++++ Positive Plate (+100 V) ++++ ---- Negative Plate (0 V) ---- 75 V 50 V 25 V Radial Field (Point Charge) 120V 60V 40V 30V + Electric Field (E) Equipotential Lines (V)

4. Examples

Example 1: Total Binding Energy of a System

Problem: Three point charges are fixed at the corners of an equilateral triangle of side length $L = 0.50\text{ m}$. The charges are $q_1 = +2.0\text{ }\mu\text{C}$, $q_2 = -3.0\text{ }\mu\text{C}$, and $q_3 = +4.0\text{ }\mu\text{C}$. Calculate the total electric potential energy of this system (the work required to assemble it from infinity).


Solution:

The total potential energy of a multi-charge system is the scalar sum of the potential energies of every unique pair of charges. There are three pairs here: $(q_1, q_2)$, $(q_2, q_3)$, and $(q_1, q_3)$.

$$E_{p, total} = \frac{k q_1 q_2}{L} + \frac{k q_2 q_3}{L} + \frac{k q_1 q_3}{L}$$ $$E_{p, total} = \frac{k}{L} (q_1 q_2 + q_2 q_3 + q_1 q_3)$$

Substitute the values (remembering to include the signs of the charges and converting $\mu\text{C}$ to $\text{C}$):

$$q_1 q_2 = (2.0 \times 10^{-6})(-3.0 \times 10^{-6}) = -6.0 \times 10^{-12}\text{ C}^2$$ $$q_2 q_3 = (-3.0 \times 10^{-6})(4.0 \times 10^{-6}) = -12.0 \times 10^{-12}\text{ C}^2$$ $$q_1 q_3 = (2.0 \times 10^{-6})(4.0 \times 10^{-6}) = +8.0 \times 10^{-12}\text{ C}^2$$

Summing the charge products:

$$\sum q_i q_j = -6.0 - 12.0 + 8.0 = -10.0 \times 10^{-12}\text{ C}^2$$

Now, calculate the total energy:

$$E_{p, total} = \frac{8.99 \times 10^9}{0.50} \times (-10.0 \times 10^{-12}) = \mathbf{-0.18\text{ J}}$$

Conclusion: The negative sign indicates that the system is bound together; an external agent would need to do $0.18\text{ J}$ of work to tear these charges apart and scatter them to infinity.

Example 2: Distance of Closest Approach (Rutherford Scattering)

Problem: An alpha particle ($q = +2e$, $m = 6.64 \times 10^{-27}\text{ kg}$) is fired head-on at a stationary gold nucleus ($Q = +79e$) with an initial kinetic energy of $5.0\text{ MeV}$. Assuming the gold nucleus remains perfectly stationary, calculate the distance of closest approach between the alpha particle and the center of the gold nucleus.

[Create image here: Alpha particle fired at a gold nucleus showing distance of closest approach d]

Solution:

Step 1: Energy Conservation
At an infinite distance, all energy is kinetic ($E_k$). At the distance of closest approach ($d$), the alpha particle momentarily stops, meaning all its kinetic energy has been converted entirely into electric potential energy ($E_p$).
$$E_{k, initial} = E_{p, final}$$ $$E_k = \frac{k q Q}{d}$$

Step 2: Convert Units
Convert $5.0\text{ MeV}$ into Joules:
$$E_k = 5.0 \times 10^6\text{ eV} \times (1.60 \times 10^{-19}\text{ J/eV}) = 8.0 \times 10^{-13}\text{ J}$$ The charges are $q = 2(1.60 \times 10^{-19})\text{ C}$ and $Q = 79(1.60 \times 10^{-19})\text{ C}$.

Step 3: Solve for $d$
$$d = \frac{k q Q}{E_k}$$ $$d = \frac{(8.99 \times 10^9)(2 \times 1.60 \times 10^{-19})(79 \times 1.60 \times 10^{-19})}{8.0 \times 10^{-13}}$$ $$d = \frac{3.64 \times 10^{-26}}{8.0 \times 10^{-13}} = \mathbf{4.55 \times 10^{-14}\text{ m}}$$

Conclusion: The alpha particle gets to within $\approx 45.5\text{ fm}$ of the gold nucleus before the electrostatic repulsion forces it to turn around.

Example 3: Potential Gradient and Kinematics

Problem: The electric potential $V$ along the x-axis in a specific region of space is given by the function $V(x) = 200x^2 - 150x$, where $V$ is in volts and $x$ is in meters. An electron is released from rest at $x = 0.50\text{ m}$. Calculate the magnitude and direction of the initial acceleration of the electron.


Solution:

Step 1: Find the Electric Field function $E(x)$
The electric field is the negative gradient (derivative) of the potential with respect to $x$: $$E(x) = -\frac{dV}{dx} = -\frac{d}{dx}(200x^2 - 150x)$$ $$E(x) = -(400x - 150) = -400x + 150$$

Step 2: Evaluate the Field at $x = 0.50\text{ m}$
$$E(0.50) = -400(0.50) + 150 = -200 + 150 = -50\text{ V m}^{-1}$$ The field is $50\text{ V m}^{-1}$ pointing in the negative x-direction.

Step 3: Calculate Force and Acceleration
The force on the electron ($q = -e$) is: $$F = qE = (-1.60 \times 10^{-19}\text{ C}) \times (-50\text{ V m}^{-1}) = +8.0 \times 10^{-18}\text{ N}$$ Notice the negative charge multiplied by the negative field results in a force in the positive x-direction.

Using Newton's Second Law ($a = F/m$) with $m_e = 9.11 \times 10^{-31}\text{ kg}$: $$a = \frac{8.0 \times 10^{-18}}{9.11 \times 10^{-31}} = \mathbf{8.78 \times 10^{12}\text{ m s}^{-2}}$$ The acceleration is $8.78 \times 10^{12}\text{ m s}^{-2}$ in the $+x$ direction.