Potential Difference Path Independence

The potential difference [math]\displaystyle{ \Delta V = V_B - V_A }[/math] between two locations A and B does not depend on the path taken between the locations.

Claimed alanghauser3

The Main Idea

The potential difference between two locations A and B does not depend on the path taken between the locations. A round trip potential difference is always zero.

Potential Difference Equations

In a uniform electric field the potential difference is equal to [math]\displaystyle{ \Delta V = -\vec{E}●\Delta \vec{l} = -(E_x●\Delta x + E_y●\Delta y + E_z●\Delta z }[/math]).

In a nonuniform electric field the potential difference is equal to [math]\displaystyle{ \textstyle\int\limits_{i}^{f}-Edl }[/math]

Examples

Simple Example of Two Different Paths

Calculate the potential difference going from A to C: [math]\displaystyle{ \Delta V = V_C - V_A =  ? }[/math]

Path 1

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Angled Path.

Since the electric field inside the capacitor is uniform all along the path we can use the equation for a uniform electric field [math]\displaystyle{ \Delta V = -\vec{E}●\Delta \vec{l} = -(E_x●\Delta x + E_y●\Delta y + E_z●\Delta z }[/math])

The displacement vector: [math]\displaystyle{ \Delta l = \lt \Delta x, \Delta y, \Delta z\gt = \lt (x_1 - 0),(-y_1 - 0)\gt = \lt (x_1,-y_1)\gt }[/math]

The electric field vector is given as: [math]\displaystyle{ \vec{E} = \lt (E_x,0,0)\gt }[/math]

Therefore the potential difference between A and C is: [math]\displaystyle{ \Delta V = -\vec{E}●\Delta \vec{l} = -E_x(x_1) + 0(-y_1) + 0(0) = -E_xx_1 }[/math]

Path 2

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Path 2

Along the path from A to B:

The displacement vector: [math]\displaystyle{ \Delta l = \lt \Delta x, \Delta y, \Delta z\gt = \lt (x_1 - 0),(0 - 0)\gt = \lt (x_1,0)\gt }[/math]

The potential difference between A and B is: [math]\displaystyle{ V_B - V_A = -\vec{E}●\Delta \vec{l} = -E_x(x_1) + 0(0) + 0(0) = -E_xx_1 }[/math]

Along the path from B to C:

The displacement vector: [math]\displaystyle{ \Delta l = \lt \Delta x, \Delta y, \Delta z\gt = \lt (x_1 - x_1),(0 - y_1)\gt = \lt (0,-y_1)\gt }[/math]

The potential difference between B and C is: [math]\displaystyle{ V_C - V_B = -\vec{E}●\Delta \vec{l} = -E_x(0) + 0(-y_1) + 0(0) = 0 }[/math]

Therefore the potential difference from A to C is: [math]\displaystyle{ \Delta V = (V_B - V_A) + (V_C - V_B) = -E_xx_1 + 0 = -E_xx_1 }[/math]

Two Different Paths Near a Point Charge

Along a straight path from a point charge Q we know that [math]\displaystyle{ \Delta V = \frac{1}{4 \pi \epsilon_0 }Q(\frac{1}{r_2} - \frac{1}{r_1}) }[/math]

Path 2

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Complicated Path

From the initial point [math]\displaystyle{ i }[/math] to point [math]\displaystyle{ A }[/math], [math]\displaystyle{ \vec{E} }[/math] is perpendicular to [math]\displaystyle{ \Delta l }[/math] so [math]\displaystyle{ \Delta V_1 = 0 }[/math]

From [math]\displaystyle{ A }[/math] to [math]\displaystyle{ B }[/math]: [math]\displaystyle{ \Delta V_2 = \frac{1}{4 \pi \epsilon_0 }Q(\frac{1}{r_3} - \frac{1}{r_1}) }[/math]

From [math]\displaystyle{ B }[/math] to [math]\displaystyle{ C }[/math]: [math]\displaystyle{ \Delta V_3 = 0 }[/math], since [math]\displaystyle{ \vec{E} }[/math] is perpendicular to [math]\displaystyle{ \Delta l }[/math].

From C to [math]\displaystyle{ f }[/math]: [math]\displaystyle{ \Delta V_4 = \frac{1}{4 \pi \epsilon_0 }Q(\frac{1}{r_2} - \frac{1}{r_3}) }[/math]

To find [math]\displaystyle{ V_f - V_i }[/math] add up all the [math]\displaystyle{ \Delta V's }[/math]

[math]\displaystyle{ V_f - V_i = \Delta V_1 + \Delta V_2 + \Delta V_3 + \Delta V_4 }[/math]

[math]\displaystyle{ = 0 + \frac{1}{4 \pi \epsilon_0 }Q(\frac{1}{r_3} - \frac{1}{r_1}) + 0 + \frac{1}{4 \pi \epsilon_0 }Q(\frac{1}{r_2} - \frac{1}{r_3}) }[/math]

[math]\displaystyle{ = \frac{1}{4 \pi \epsilon_0 }Q(\frac{1}{r_2} - \frac{1}{r_1}) }[/math]

Round Trip Potential Difference

Since the only points that matter when calculating the potential difference are the initial and final locations, then the round trip potential of any path will equal [math]\displaystyle{ 0 }[/math] since the initial and final locations are the same.

Take for example this very simple circuit:

Simple Circuit

The potential produced by the battery is equivalent to its [math]\displaystyle{ emf }[/math].

Since circuits obey the conservation of energy the potential difference across the whole circuit must equal [math]\displaystyle{ 0 }[/math]

Therefore: [math]\displaystyle{ \Delta V = V_b + V_w = emf + (-EL) = 0 }[/math]

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