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
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
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
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]
Difficult
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