Where can I find the image charge for a point charge near conducting surfaces?

In summary, two semi-infinite grounded conducting planes meet at right angles. Between them is a point charge q at a distance of 40cm. If the image charge of the point charge is located at (-40,40), (-40,-40), (40,-40) then the point charge has two images. One image is located at (-40,40), the other is located at (-40,-40).
  • #1
Reshma
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Someone's got to help me on these two.

1)Two semi-infinite grounded conducting planes meet at right angles. In the region between them, a point charge q is kept such that it is at a distance of 40cm from each of the two planes[i.e.(40,40)]. Find the position of its image charge/charges.

2)A point charge q is placed in the vicinity of a conducting sphere. Find the poition of its image charge.

Just in case...An image charge will produce zero potential together with q.
 
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  • #2
Reshma said:
Someone's got to help me on these two.

1)Two semi-infinite grounded conducting planes meet at right angles. In the region between them, a point charge q is kept such that it is at a distance of 40cm from each of the two planes[i.e.(40,40)]. Find the position of its image charge/charges.

2)A point charge q is placed in the vicinity of a conducting sphere. Find the poition of its image charge.

Just in case...An image charge will produce zero potential together with q.
Have you thought about what the optical image(s) would be if you placed an object in the symmetry plane between two mirrors at right angles? How many images would be formed, and where would they be? For charges, image charges must have the opposite sign to have constant (zero) potential at the surface. When multiples surfaces are involved there can be images of images, just as there are for optical images.
 
  • #3
OlderDan said:
Have you thought about what the optical image(s) would be if you placed an object in the symmetry plane between two mirrors at right angles? How many images would be formed, and where would they be? For charges, image charges must have the opposite sign to have constant (zero) potential at the surface. When multiples surfaces are involved there can be images of images, just as there are for optical images.

Thank you very much! I think this solves my first problem.

Since my charge is located in the 1st quadrant, I should have one image charge in each of the other three quadrants(total=3). So by your explanation my image charges are located at (-40,40), (-40,-40), (40,-40). Is my answer correct?

However in the 2nd problem, I don't know how to exploit the symmetry for the sphere. Can you help me on that?
 
  • #4
Can someone help?
 

1. What is the concept behind locating image charges?

The concept behind locating image charges is to create a mirror image of a given charge at a specific location in order to simulate the electric field and potential at that location.

2. How do you determine the location of an image charge?

The location of an image charge can be determined using the principle of mirror charges, which states that the distance between the original charge and its image charge should be equal to the distance between the image charge and the reflecting surface.

3. Can image charges only be used for point charges?

No, image charges can also be used for other charge distributions, such as line charges and surface charges. The concept of mirror charges still applies, but the calculation becomes more complex.

4. What are the limitations of using image charges?

One limitation of using image charges is that it only works for simple geometries and symmetrical charge distributions. It also assumes that the conducting surface is perfectly conducting and has no thickness.

5. How is the electric potential calculated using image charges?

The electric potential at a point due to an image charge is calculated by summing the potentials from the original charge and its image charge using the principle of superposition. The potential at the point due to the image charge is equal in magnitude but opposite in sign to the potential due to the original charge.

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