Find potential inside spherical shell

AI Thread Summary
The discussion focuses on calculating the electric potential at point A, located R/2 from the center of a conducting sphere with charge Q, influenced by an external charge q placed at a distance of 2R. The initial approach incorrectly sums the potentials from the sphere and charge q, suggesting V = kQ/R and V = 2kq/5R. However, it is noted that the sphere is an equipotential region, and the influence of charge q complicates the distribution. The correct method involves using the concept of image charges to account for the non-symmetrical charge distribution. Ultimately, the potential at the center due to the sphere remains kQ/R, while the potential from charge q must also be considered to find the total potential at point A.
utkarshakash
Gold Member
Messages
852
Reaction score
13

Homework Statement


A conducting sphere of radius R has a charge Q. A particle carrying a charge q is placed a distance 2R from the sphere. Find the potential at point A located a distance R/2 from the center of the sphere on the line connecting the center of the sphere and particle q. Note that the charge distribution of the sphere is not symmetrical due to the influence of particle q


The Attempt at a Solution



The potential at point A will be equal to sum of potential due to sphere and charge 'q'.
Due to sphere, V= kQ/R
and due to charge, V= 2kq/5R

But this is not the correct answer.
 

Attachments

  • image018.gif
    image018.gif
    3.6 KB · Views: 581
Physics news on Phys.org
utkarshakash said:
The potential at point A will be equal to sum of potential due to sphere and charge 'q'.
Due to sphere, V= kQ/R
and due to charge, V= 2kq/5R

But this is not the correct answer.

The item in red is not correct .
 
utkarshakash said:

Homework Statement


A conducting sphere of radius R has a charge Q. A particle carrying a charge q is placed a distance 2R from the sphere. Find the potential at point A located a distance R/2 from the center of the sphere on the line connecting the center of the sphere and particle q. Note that the charge distribution of the sphere is not symmetrical due to the influence of particle q


The Attempt at a Solution



The potential at point A will be equal to sum of potential due to sphere and charge 'q'.
Due to sphere, V= kQ/R
and due to charge, V= 2kq/5R

But this is not the correct answer.

Hint:the whole sphere is an equipotential region.
 
utkarshakash said:

Homework Statement


A conducting sphere of radius R has a charge Q. A particle carrying a charge q is placed a distance 2R from the sphere. Find the potential at point A located a distance R/2 from the center of the sphere on the line connecting the center of the sphere and particle q. Note that the charge distribution of the sphere is not symmetrical due to the influence of particle q .


The Attempt at a Solution



The potential at point A will be equal to sum of potential due to sphere and charge 'q'.
Due to sphere, V= kQ/R
and due to charge, V= 2kq/5R

But this is not the correct answer.
The reason that V= kQ/R is incorrect for the sphere is highlighted in RED above.

Use the concept of image charges .
 
projjal said:
Hint:the whole sphere is an equipotential region.

Will the potential at centre be equal to kQ/R due to sphere?
 
utkarshakash said:
Will the potential at centre be equal to kQ/R due to sphere?

Yeah.

You also know the potential at centre due to charge q and wid that you get the answer.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top