Electric Field Of Charged Disk

Electric Field Of Charged Disk

Edited By Vishal kumar | Updated on Sep 26, 2024 11:50 AM IST

The electric field of a uniformly charged disk is a fundamental problem in electrostatics, frequently encountered in physics. A charged disk generates an electric field that varies with distance from its surface, and its distribution of charge plays a crucial role in determining the field's characteristics. The field is typically analyzed in the plane perpendicular to the disk's surface, with special attention to the axis passing through its centre.

Electric Field Due to Uniformly Charged Disk

Let us take a disk of radius R with a uniform positive surface charge density (charge per unit area) σ. Our aim is to find an electric field at a point on the axis of the disk at a distance x from its centre.

From the figure, we can see that we have taken a typical ring that has charge dQ, inner radius r and outer radius r+dr. Its area dA

dEx=14πε0(2πσrdr)x(x2+r2)3/2
If we integrate from 0 to R, we will get the total field -

Ex=dEx=0RdEx=0R14πε0(2πσrdr)x(x2+r2)3/2
Here, ' x ' is constant and 'r' is the variable. After integration, we get -

Ex=σx2ε0[1x2+R2+1x]=σ2ε0[1xx2+R2]
As this disc is symmetric to the x -axis, the field in the rest of the component is zero i.e., Ey=Ez=0

Special case -
1) When R>>x, then Ex=σ2ε0 Note that this equation is independent of ' x '
2) When x0 (i.e very near to the disc), then Ex=σ2ε0

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Some Solved Examples:

Example 1: A thin disc of radius b=2a has a concentric hole of radius 'a' in it (see figure). It carries a uniform surface charge '\sigma 'on it. If the electric field on its axis at height' h ' (h<<a) from its centre is given as 'Ch' then value of 'C' is :


σσaϵ0σ2)2aϵ0

3) σ4aϵ0

σ8)8aϵ0
Solution:

As we discussed in

Uniformly charged disc -

E=σ2ϵ0[1x(x2+R2)12]V=σ2ϵ0[x2+R2x]
- wherein

Electric Field due to complete disc (R=2a) at distance x

E1=σ2ϵo[1x(R2+x2)12][h=x;2a=R]E1=σ2ϵo[1h(4a2+h2)12]=σ2ϵo[1h2a]
Similarly, the electric field due to disc (R=a)

E2=σ2ϵo[1ha]
Now E=E1E2=σ2ϵo[1h2a]σ2ϵo[1ha]=σh4ϵoa
Hence C=σ4ϵoa

Example 2: What will be the electric field due to a uniformly charged disc At a distance x from centre O on its axis if X0

E=σ4ε0Eσ2ε0E=2KλR=Q2π2ε0R2

4)0

Solution:
As we learned
Uniformly charged disc -

If x0Eσ2ϵ0

wherein

i.e. Point situated near the disc it behaves as an infinite sheet of charge.

E=σ2ε0[1xx2+R2] Putx =0

Example 3 The surface charge density of a thin charged disc of radius R is \sigma . The value of the electric field at the centre of the disc is σ2ϵ0y With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc :
1)reduces by 70.7%
2)reduces by 29.3%
3)reduces by 9.7%
4)reduces by 14.6%

Solution:
Electric field intensity at the centre of the disc

E=σ2ϵ0( given )
Electric field along the axis at any distance x from the centre of the disc

E=σ2ϵ0(1xx2R2)
From question, x=R (radius of disc)

E=σ2ϵ0(1RR2+R2)=σ2ϵ0(2RR2R)=414E
So reduction in the value of electric field

=(E414E)×100E=100014%=70.7%

Example 4:Find out the surface charge density at the intersection of point x=3 m plane and the x -axis in the region of uniform line charge of 8nC/m along the z -axis in free space.
1) 47.88C/m
2) 0.424nCm2
3) 0.07nCm2
4) 4.0nCm2

Solution:
Electric field due to uniformly charged rod- E=2Kλr

Electric field due to uniformly charged disk-

E=σ2ε02 Kλr=σε0σ=2Kλε0r=λ2πr=8×1092π×3σ=0.424×109Cm2

Summary:

To compute the electric field at a point along the axis of a uniformly charged disk, the disk is treated as a series of infinitesimal rings of charge. Each ring contributes to the net electric field, with its components integrated to find the total field. For a point close to the disk, the field behaves like that of a charged plane, while at far distances, it resembles the field of a point charge. The formula derived depends on the radius of the disk, surface charge density, and the distance from the point of interest to the disk’s centre.

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