NEET Class neet Answered
HELP
![question image](https://images.topperlearning.com/topper/new-ate/topr_14936710337527135783A24D2724A84E2AB9E1378689CCC71F.jpeg)
Asked by astutijoshi | 28 Jul, 2019, 16:01: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/b91030bb654f7403df390128f274da985d3e55e523ebb4.56557131ring.png)
Fig.1 shows a charged ring of radius R that has charge density λ per unit length.
We need to find elecrical intensity at a point P which is on the axis passing through centre of ring and at a distance z from the centre.
we consider a small length dl in the ring that has charge λdl. Direction of electric field dE is shown in figure.
![begin mathsize 12px style d E space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction 1 over s squared space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction fraction numerator 1 over denominator open parentheses R squared space plus space z squared close parentheses end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/9178bbfc18a539fd87b0e1977d0b3b50.png)
Electric field dE can be resolved in axial direction as dEz and radial direction dEr as shown infigure.
Axial field : ![begin mathsize 12px style d E subscript z space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction 1 over s squared space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction fraction numerator 1 over denominator open parentheses R squared space plus space z squared close parentheses end fraction space cos theta space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction z over open parentheses R squared space plus space z squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end style](https://images.topperlearning.com/topper/tinymce/cache/6203bb0ee9febd777106cbcc45e45af8.png)
![begin mathsize 12px style d E subscript z space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction 1 over s squared space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction fraction numerator 1 over denominator open parentheses R squared space plus space z squared close parentheses end fraction space cos theta space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction z over open parentheses R squared space plus space z squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end style](https://images.topperlearning.com/topper/tinymce/cache/6203bb0ee9febd777106cbcc45e45af8.png)
When we add the electric field due to whole ring by integration, Radial component will be cancelled out.
Hence Electric field of whole charged ring at a point P =
..................(2)
![begin mathsize 12px style integral d E subscript z space equals space space integral fraction numerator lambda d l over denominator 4 pi epsilon subscript o end fraction z over open parentheses R squared space plus space z squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent space equals space fraction numerator 2 pi R space lambda over denominator 4 pi epsilon subscript o end fraction z over open parentheses R squared space plus space z squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end style](https://images.topperlearning.com/topper/tinymce/cache/a9a5b0f53249f93641357d08aeddb53a.png)
In the above eqn.(2), 2πRλ in the numerator is the total charge on the ring.
Now let us calculate electric field due to a circular disc of radius R that has charge density σ per unit area.
Fig.2 shows such a disc. Let us consider a ring which is part of disc that has radius r, thickness dr as shown in figure.
Using eqn.(2), we get the electric field dE due to this ring at a point P which is at a distance z from centre of disc is given by
![begin mathsize 12px style d E space equals fraction numerator open parentheses 2 pi r space d r close parentheses space sigma over denominator 4 pi epsilon subscript o end fraction z over open parentheses r squared space plus space z squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end style](https://images.topperlearning.com/topper/tinymce/cache/b90d95ab245e560aff8c88da9dcc95b9.png)
Electric field due to charged disc is obtained by integrating above eqn.(3).
![begin mathsize 12px style E space equals space integral d E space equals fraction numerator sigma space z over denominator 4 space epsilon subscript o end fraction integral subscript 0 superscript R fraction numerator 2 r space d r over denominator open parentheses r squared space plus space z squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end fraction space equals space fraction numerator sigma space over denominator 2 space epsilon subscript o end fraction open square brackets 1 space minus space fraction numerator z over denominator square root of R squared plus z squared end root end fraction close square brackets end style](https://images.topperlearning.com/topper/tinymce/cache/f241a51da33f5ec9a53f260d784ee782.png)
From eqn.(4), by substituting σ = 10 nC/m2 , R = 4 cm and z = 3 cm, we get electric field E = 0.226×103 N/C
Answered by Thiyagarajan K | 29 Jul, 2019, 07:43: AM
Application Videos
Concept Videos
NEET neet - Physics
Asked by cmashudrana | 26 Jul, 2024, 10:51: AM
NEET neet - Physics
Asked by upparmanjunath70 | 19 Jul, 2024, 22:26: PM
NEET neet - Physics
Asked by zfaima704 | 19 Jul, 2024, 14:29: PM
NEET neet - Physics
Asked by shaikmuneerjaan | 18 Jul, 2024, 20:58: PM
NEET neet - Physics
Asked by nk4746870 | 15 Jul, 2024, 21:21: PM
NEET neet - Physics
Asked by dasmyton | 09 Jul, 2024, 22:46: PM
NEET neet - Physics
Asked by shobhaguduru02 | 09 Jul, 2024, 20:21: PM
NEET neet - Physics
Asked by kashinathpaikrao89446 | 09 Jul, 2024, 10:46: AM
NEET neet - Physics
Asked by dshyamala44 | 08 Jul, 2024, 21:05: PM
NEET neet - Physics
Asked by jagadishshnkr | 08 Jul, 2024, 20:50: PM