10.5 Electric Potential

Electric Potential Due to Charged Objects

Electric potential represents the electric potential energy per unit charge at a specific point in space. This scalar quantity allows us to analyze how charged particles interact with electric fields without directly considering forces.

When multiple charges are present, we can use scalar superposition to find the total electric potential. Unlike electric fields (which are vectors), electric potentials simply add algebraically:

$V=\frac{1}{4 \pi \varepsilon_{0}} \sum_{i} \frac{q_{i}}{r_{i}}$

This equation shows that the electric potential:

• Increases with larger source charges ($q_i$)
• Decreases with distance ($r_i$)
• Can be positive or negative depending on the sign of the charges

Electric potential difference between two points equals the change in electric potential energy per unit charge when moving a test charge between those points:

$\Delta V=\frac{\Delta U_{E}}{q}$

This potential difference can arise from:

• Spatial separation in an electric field
• Chemical processes that separate charge (as in batteries)
• Other mechanisms that create charge separation

An important property of conductors is that when they are in electrical contact, electrons will redistribute until all connected conducting surfaces reach the same electric potential. This is why connected conductors in electrostatic equilibrium have uniform potential across their surfaces.

Relationship Between Electric Potential and Electric Field

The electric field and electric potential are intimately related. The average electric field between two points equals the electric potential difference divided by the distance between them:

$||\vec{E}||=\left|\frac{\Delta V}{\Delta r}\right|$

This relationship reveals several important principles:

1. Electric field vectors point in the direction of decreasing potential (from high to low potential)
2. The magnitude of the electric field equals the rate of change of potential with distance

To visualize these concepts, we use two complementary representations:

• Electric field vector maps show the direction and strength of the electric field at various points
• Equipotential lines (isolines) connect points of equal electric potential

These representations have specific properties:

• Equipotential lines are always perpendicular to electric field vectors
• No work is done when moving a charge along an equipotential line
• The closer together equipotential lines are, the stronger the electric field
• An isoline map can be constructed from an electric field vector map, and vice versa

Practice Problem 1: Electric Potential Due to Multiple Charges

Solution

To solve this problem, we need to apply the principle of scalar superposition, calculating the contribution from each charge and then adding them together.

Step 1: Identify the position of each charge and the point where we need to find the potential.

• Charge 1: q₁ = +2.0 μC at (0, 0) m
• Charge 2: q₂ = -3.0 μC at (2, 0) m
• Charge 3: q₃ = +1.0 μC at (4, 0) m
• Point P is at (3, 0) m

Step 2: Calculate the distance from each charge to point P.

• r₁ = |3 - 0| = 3.0 m
• r₂ = |3 - 2| = 1.0 m
• r₃ = |3 - 4| = 1.0 m

Step 3: Calculate the electric potential due to each charge using V = k(q/r).

• V₁ = (9 × 10⁹ N·m²/C²)(2.0 × 10⁻⁶ C)/(3.0 m) = 6,000 V
• V₂ = (9 × 10⁹ N·m²/C²)(-3.0 × 10⁻⁶ C)/(1.0 m) = -27,000 V
• V₃ = (9 × 10⁹ N·m²/C²)(1.0 × 10⁻⁶ C)/(1.0 m) = 9,000 V

Step 4: Add the individual potentials to find the total potential at point P.

• V_total = V₁ + V₂ + V₃ = 6,000 V + (-27,000 V) + 9,000 V = -12,000 V

Therefore, the electric potential at point (3.0 m, 0 m) is -12,000 V or -12 kV.

Practice Problem 2: Equipotential Lines and Electric Field

Solution

In a uniform electric field, the electric potential varies only in the direction of the field. Since the field points in the positive x-direction, the potential will change only with the x-coordinate.

Step 1: Use the relationship between electric field and potential difference. Since E = -ΔV/Δx, we can rearrange to find ΔV = -E·Δx

Step 2: Calculate the potential at the point (3.0 m, 4.0 m). The x-coordinate is 3.0 m, and the field is 200 N/C in the positive x-direction. V = V₀ - E·x = 0 - (200 N/C)(3.0 m) = -600 V

Step 3: Sketch the equipotential line. The equipotential line passing through (3.0 m, 4.0 m) would be a vertical line at x = 3.0 m, since all points with x = 3.0 m have the same potential of -600 V, regardless of their y-coordinate. This line would be perpendicular to the electric field vectors, which all point in the positive x-direction.

Therefore, the electric potential at the point (3.0 m, 4.0 m) is -600 V, and the equipotential line is a vertical line at x = 3.0 m.