Electric Field due to Infinite Sheet of Charge

Electric Field due to Infinite Sheet of Charge

AU : May-03, 06, 07, 09, 10, Dec.-03,04, 06, 07, 09, 10, 16 June-09

• Consider an infinite sheet of charge having uniform charge density ρ₅ C/m², placed in xy plane as shown in the Fig. 2.8.1. Let us use cylindrical coordinates.

• The point P at which  to be calculated is on z-axis.

• Consider the differential surface area dS carrying a charge dQ. The normal direction to dS is z direction hence dS normal to z direction is r dr dϕ.

Now dQ = ρ_S dS = p S r dr dϕ ... (2.8.1)

• The distance vector   has two components as shown in the Fig. 2.8.2.

• For infinite sheet in xy plane, r varies from 0 to while varies from 0 to 2π.

Note : As there is symmetry about z-axis from all radial direction, all  components of   are going to cancel each other and net  will not have any radial component.

• Hence while integrating d there is no need to consider  component. Though if considered, after integration procedure, it will get mathematically cancelled.

... For points above xy plane

• Now  is direction normal to differential surface area dS considered. Hence in general if  is direction normal to the surface containing charge, the above result can be generalized as,

... For points below xy plane.

Note : The equation (2.8.6) is standard result and can be used directly to solve the problems.

Key Point : Thus electric field due to infinite sheet of charge is everywhere normal to the surface and its magnitude is independent of the distance of a point from the plane containing the sheet of charge.

Important observations :

1.   due to infinite sheet of charge at a point is not dependent on the distance of that point from the plane containing the charge.

2. The direction of   is perpendicular to the infinite charge plane.

3. The magnitude of   is constant every where and given by  

Ex. 2.8.1 Find the force on a point charge q located at (0,0, h) m due to charge of surface charge density ρ_s C/m² uniformly distributed over the circular disc r ρ ≤ a, z = 0 m Also find electric field intensity at the same point.AU : May-03, 06, 07, 09, 10, Dec.-03, 04, 07, 09, 10, 16, Marks 10

Sol. : The charges are shown in the Fig. 2.8.4.

Consider the differential area dS carrying the charge dQ. The normal direction to dS is  hence

dS_z = r dr dϕ

dQ = ρ_S dS = ρ_S r dr dϕ

Thus the force on a point charge q due to dQ is,

The    can be splitted as shown in the Fig. 2.8.4 (b),

Key Point : Due to symmetry about z-axis, all radial components will cancel each other. Hence there will not be any component of  So in the integration  need not be considered.

Ex. 2.8.2 A sheet of charge lies in yz plane at x = 0 and has uniform surface charge density of 5.0 pC/m2. Find the electric field at a point P (- 5, 0, 0) on x-axis. AU: Dec.-06, June-09, Marks 10

Sol. : The sheet is shown in the Fig. 2.8.5.

The point P is on the back side of the plane.

The Normal to the plane in the direction of P is  

Ex. 2.8.3 An infinite sheet with surface charge Q = 12 ε₀ Cm^-2 is lying in the plane x - 2y + 3z = 4. Find an expression for the field-intensity on the side of the plane containing the origin.

Sol. : The plane is shown in the Fig. 2.8.6. The plane can be defined uniquely from three points which can be obtained from the equation of plane x-2y + 3z = 4.

For x = 0 , y = 0 , z = 4/3 P(0,0,4/3)

For x = 0, z = 0, y = -2  Q(0.-2,0)

For y = 0, z = 0, x = 4 (4,0,0)

The three poionts P, Q and R define a plane.

 The plane is infinite sheet of charge.

Note : If plane is defined as Ax + By + Cz = D then the unit vector normal to the plane is,

Positive sign for front side of the plane and negative sign for back side of the plane.

In this case, A = 1, B = - 2, C = 3, D = 4

The origin is on the back side of the plane so use negative sign.

Examples for Practices

Ex. 2.8.4 Charge lies in y = - 5 m plane in the form of an infinite square sheet with a uniform charge density of ps = 20 nC/m2. Determine   at all the points.    

Ex. 2.8.5 Determine the force on a point charge of 5 nC at (0, 0, 5) m due to uniformly distributed charge of 5 mC over a circular disc of radius r ≤  lm in z = 0 plane.

Ex. 2.8.6 Find   at P (1, 5, 2) m in free space if a point charge of 6 uC is located at (0,0, 1), the uniform line charge density

PL = 180 nC/m along ×  axis and uniform sheet of charge with ps = 25 nC/m² over the plane z = -1.

Ex. 2.8.7 The charge lies on the circular disc r ≤ 4 m, z = 0,   with density ps = [l0-4/r] c/m2.

Determine   at r = 0, z = 3 m.

Ex. 2.8.8 A sheet of charge lies in yz plane at x = 0 and has uniform surface charge density of 5.0 pC/m2. Find the electric field at a point P (- 5, 0, 0) on x-axis.

Ex. 2.8.9 A circular disc of 10 cm radius is charged uniformly with a total charge. Find   at a point 20 cm on its axis.

Ex. 2.8.10 A charge distribution is placed in the z = -3 m plane in the form of a square sheet defined by -2 ≤  x ≤  2m, -2 ≤ y ≤ 2m. It has a charge density of ps = 2(x² + y² + 9)^3/2 nC/m². Find the electric field intensity () at the origin.

Ex. 2.8.11 Three infinite uniform sheets of charge are located in free space follows 3 nC/m² at z = - 4, 6 nC/m² at z = 1 and - 8 nC/m² at z = 4.

Find   at the point :

i) P_A = (2, 5, -5),

ii) P_B = (4, 2, -3),

iii) P_C = (-1, -5, 2),

iv) P_D = (-2, 4, 5).

Ex. 2.8.12 Find electric field intensity ( ) at origin if the following charge distributions are present in the free space :

i) Point charge 12 nC at P(2, 0, 6),

ii) Uniform line charge density 3 nC/m at x = - 2, y =3.

iii) Uniform surface charge density 0.2 nC/m2 at x = 2.

Review Questions

1. Find   due to infinite sheet of charge placed in x-y plane and having uniform surface charge density of ps C/m² .

2. A circular disc of radius 'a' m is charged uniformly with a charge density of ps C/m². Find the electric field at a point 'h' m from the disc along its axis.