So, why is the potential decreasing as we move away from the origin?

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The discussion centers on the decrease in electric potential as one moves away from the origin, where the potential is initially 10V and reduces to 8V after moving one unit along any axis. The potential continues to decrease, leading to a calculated potential of 4V at the point (1,1,1). Participants seek a mathematical explanation for this phenomenon, noting that the electric field is uniform and the potential decreases consistently with distance. Suggestions include using geometric and mathematical methods to analyze the relationship between potential and distance. The conversation emphasizes the need for clarity in understanding how electric potential changes in a uniform field.
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Hi friends Problem is -

https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash3/s480x480/548739_2677144185130_1329975308_n.jpg

Attempt -

Well the potential at the origin is 10V. But if we move a distance of 1unit from the origin in any axis, the potential becomes 8 V. The difference is only 2V. The potential is decreasing as we move away from the origin. And the point (1,1,1) is not too far from point (1,0,0). Hence the potential decreases by a small amount It'll not be zero because otherwise the difference would be of 8 V, which is unsatisfactory. Hence the answer would be 4 V.

And the answer is also correct according to the book.
But friends the problem is that I am unable to make any mathematical explanation for this question. Please fiends give your suggestions.

Thank you all in advance.
 
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Geometric method: If you draw the projection of the cube onto a piece of paper such that the 8V plane formed by the points (001), (010), and (100) is viewed "on edge" (and the plane formed by (110), (101), and (011) also viewed on edge), you will see the line segment connecting (000) and (111), normal to these planes, is divided equally into thirds.

Math method: Find the [shortest] distance between (000) and the 8V plane. Find the distance between (000) and (111). Proportions.
 
Determine the E field from the given potentials using ΔV = -∫E*ds from the origin to each of x, y and z positions, where ΔV = -2V for each direction,

where ds = dx i + dy j + dz k.

Then integrate -∫E*ds along each of the three directions to get ΔV from (0,0,0) to (1,1,1).

Finally of course answer = ΔV + the voltage at (0,0,0).

Vectors are in bold face. * means dot product.
 
thunderhadron said:
Hi friends Problem is -

https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash3/s480x480/548739_2677144185130_1329975308_n.jpg

Attempt -

Well the potential at the origin is 10V. But if we move a distance of 1unit from the origin in any axis, the potential becomes 8 V. The difference is only 2V. The potential is decreasing as we move away from the origin. And the point (1,1,1) is not too far from point (1,0,0). Hence the potential decreases by a small amount It'll not be zero because otherwise the difference would be of 8 V, which is unsatisfactory. Hence the answer would be 4 V.

And the answer is also correct according to the book.
But friends the problem is that I am unable to make any mathematical explanation for this question. Please fiends give your suggestions.

Thank you all in advance.
Well, it's a uniform electric field.

If you move 1 unit in the positive x direction, the electric potential decreases by 2 Volts, no matter the starting location.

Similar results hold for moving 1 unit parallel to either of the axes, y or z.
 
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