Field at a distance caused by a ring charge

In summary, the electric field at (a) 1.2 cm, (b) 3.6 cm and (c) 4.0 m from the center of the ring is 1.54 x 10^3 N/C.
  • #1
chopnhack
53
3

Homework Statement


2.75 mC is uniformly distributed on a ring of radius 8.5 cm. Find the electric field on the axis at (a) 1.2
cm, (b) 3.6 cm and (c) 4.0 m from the center of the ring.

Homework Equations


[/B]
E = (kQx)/(a^2+x^2)^(3/2)

3. The Attempt at a Solution


This question really boils down to a calculator issue, operator error issue or the abyss of unknown. When I perform the calculation on two different calculators using different methods, including a graphing calculator that has memory allowing me to repeat an equation previously used and change its variables, I get a strange answer for the second part 3.6cm. It ends up larger by one order! It can't be because its further from the charge, but each time I calculate it I get an erroneous value.

a) 4.695 x 10^5 N/C
b) 1.13 x 10^6 N/C
c) 1.54 x 10^3 N/C

Please don't say throw out the calculator because I checked it with a 30 year old Sharp 531D and I get the same results! I must be missing something. Anyone able to assist? Most appreciated, thanks!

20170710_174008.jpg
 
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  • #2
chopnhack said:
I get a strange answer for the second part 3.6cm. It ends up larger by one order! It can't be because its further from the charge, but each time I calculate it I get an erroneous value.
Are you sure it can't be? What is the field at x = 0?
 
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  • #3
TSny said:
Are you sure it can't be? What is the field at x = 0?
Fair point, the value at x = 0 is zero, but I think that is almost like the argument of no charge inside a hollow sphere. In this case, the center is the ring, it being just a slice of a sphere.

I am not sure it can't be, but it doesn't seem logical in the sense that as you get further from the field you should see a diminished field, as is the case for all other charges we have dealt with so far.

Have you tried the calculation?
 
  • #4
chopnhack said:
I am not sure it can't be, but it doesn't seem logical in the sense that as you get further from the field you should see a diminished field, as is the case for all other charges we have dealt with so far.
Only the x-component of the field survives. Suppose you had a single point charge located on the y-axis at say, y = .30 m. Now consider different points on the x axis, say x = 0, x = .10 m, x = .20 m, ..., x = 1.0 m. Can you see how the x-component of the field of the point charge varies for these points?

Have you tried the calculation?
Yes, I get your results. Since you have a graphing calculator, you might try plotting your function E(x).
 
  • #5
TSny said:
Only the x-component of the field survives.
Yes, I get your results. Since you have a graphing calculator, you might try plotting your function E(x).

Yes, the opposing sine theta portions cancel each other leaving just the cosine portion.
I will give that a go and see if I can get it to display a graph.

That other bit you mentioned on a point charge at y - I can somewhat see what you mean, due to the circular nature of the ring, but I want to work it out later when I have some time.
 

Related to Field at a distance caused by a ring charge

1. What is a "field at a distance caused by a ring charge"?

The "field at a distance caused by a ring charge" refers to the electric or magnetic field that is produced by a charged ring and extends outwards from the ring into the surrounding space.

2. How is the field at a distance calculated for a ring charge?

The field at a distance from a ring charge can be calculated using the electric or magnetic field equations, depending on whether the ring is charged with electricity or magnetism. The equations take into account the charge or current of the ring, the distance from the ring, and other factors such as the direction and shape of the ring.

3. What factors affect the strength of the field at a distance from a ring charge?

The strength of the field at a distance from a ring charge is affected by several factors, including the amount of charge or current in the ring, the distance from the ring, and the shape and direction of the ring. Additionally, the presence of other nearby charges or magnets can also influence the strength of the field.

4. How does the field at a distance from a ring charge change as the distance increases?

The field at a distance from a ring charge follows an inverse square law, meaning that as the distance from the ring increases, the strength of the field decreases at a rate proportional to the square of the distance. This means that the field becomes weaker the farther away you are from the ring charge.

5. What are some real-world applications of the field at a distance caused by a ring charge?

The field at a distance caused by a ring charge has many practical applications, including in technologies such as electromagnets, electric motors, and generators. It is also important in understanding the behavior of charged particles in particle accelerators and in the study of magnetic fields in space, such as those produced by planets and stars.

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