Griffiths 2.26 E&M Potential Difference Cone

In summary, it is important to carefully review the calculations and make sure they are accurate and consistent in order to obtain the correct potential difference between the vertex and the center of the top of a cone.
  • #1
jad22
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This problem is finding the potential difference between the vertex (a) and the center of the top (b), for a cone with height and radius H. Can this problem be done by: finding the potential at (b) using a flat disc and then finding the potential at (a) using the surface area of a cone (without the disc). When I do this I get a potential at b = 2(pi)*sigma*h. I then find the potential at the vertex = (pi)*sigma*H*ln(1+sqrt(2)). This solution is almost exact as the one I have seen online, except my potential for the vertex is a factor of 2 off. I have used Gaussian units.
 
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  • #2


I can confirm that the method described in the forum post is a valid approach for finding the potential difference between the vertex and the center of the top of a cone. However, there may be some errors in the calculations that could be causing the discrepancy in the final result.

Firstly, it is important to note that the potential at a point is a scalar quantity and does not have units, so using Gaussian units should not affect the final result. However, if the formula for potential used in the calculation is in Gaussian units, then the values for the constants such as sigma and h should also be in Gaussian units to ensure consistency.

Secondly, it is possible that the formula used for finding the potential at the vertex may be incorrect. It is always a good practice to double-check the formula and make sure it is applicable to the specific problem at hand.

Lastly, there may be a mistake in the calculation itself, such as a misplaced decimal point or a missing factor. It is important to carefully review each step of the calculation to identify any potential errors.

In conclusion, while the method described in the forum post is valid, it is crucial to ensure that the calculations are accurate and consistent to obtain the correct result. Double-checking the formulas, using consistent units, and carefully reviewing the calculation steps can help in identifying and correcting any errors.
 

1. What is the Griffiths 2.26 E&M Potential Difference Cone?

The Griffiths 2.26 E&M Potential Difference Cone is a concept in electromagnetism that describes the change in electrical potential as a function of distance from a point charge. It is commonly used to visualize the electric field around a point charge in three-dimensional space.

2. How is the Griffiths 2.26 E&M Potential Difference Cone calculated?

The Griffiths 2.26 E&M Potential Difference Cone can be calculated using the equation V = k(Q/r), where V is the potential difference, k is the Coulomb constant, Q is the charge of the point charge, and r is the distance from the point charge.

3. What does the shape of the Griffiths 2.26 E&M Potential Difference Cone represent?

The shape of the Griffiths 2.26 E&M Potential Difference Cone represents the strength and direction of the electric field around a point charge. The cone becomes narrower and steeper as the distance from the point charge increases, indicating a stronger electric field.

4. How does the Griffiths 2.26 E&M Potential Difference Cone relate to electric potential energy?

The Griffiths 2.26 E&M Potential Difference Cone is directly related to electric potential energy. As the potential difference increases, the potential energy of a charged particle also increases. The shape of the cone can be used to determine the potential energy of a charged particle at different distances from the point charge.

5. Can the Griffiths 2.26 E&M Potential Difference Cone be applied to other types of charges?

Yes, the Griffiths 2.26 E&M Potential Difference Cone can be applied to any type of point charge, whether it is positive or negative. The only difference is the direction of the electric field lines, which will point away from a positive charge and towards a negative charge.

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