ELECTROSTATICS

Gauss’s Law and Applications

Though Coulomb’s law is fundamental, one finds it cumbersometo use it to cal-culate electric field due to a continuous charge distribution because the integralsinvolved can be quite difficult. An alternative but completely equivalent formula-tion is Gauss’s Law which is very useful in situations which exhibit certain sym-metry.

Electric Lines of Force :

Electric lines of force (also known asfield lines) is a pictorial representation of theelectric field. These consist of directed lines indicating the direction of electricfield at various points in space.

• There is no rule as to how many lines are to be shown. However, it iscustomary to draw number of lines proportional to the charge. Thus if Nnumber of lines are drawn from or into a chargeQ, 2N number of lineswould be drawn for charge2Q.

• The electric field at a point is directed along the tangent to the field lines.A positive charge at this point will move along the tangent ina directionindicated by the arrow.

• Lines are dense close to a source of the electric field and become sparse asone moves away.

• Lines originate from a positive charge and end either on a negative chargeor move to infinity.

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• Lines of force due to a solitary negative charge is assumed tostart at infinityand end at the negative charge.

• Field lines do not cross each other. ( if they did, the field at the point ofcrossing will not be uniquely defined.)

• A neutral point is a point at which field strength is zero. Thisoccurs becauseof cancellation of electric field at such a point due to multiple charges.

Exercise : Draw field lines and show the neutral point for a charge+4Q locatedat (1, 0) and−Q located at(−1, 0).

2.3 Electric Flux

The concept offlux is borrowed from flow of water through a surface. The amountof water flowing through a surface depends on the velocity of water, the area of thesurface and the orientation of the surface with respect to the direction of velocityof water.Though an area is generally considered as a scalar, an element of area may beconsidered to be a vector because :

• It has magnitude (measured in m2).

• If the area is infinitisimally small, it can be considered to be in a plane. Wecan then associate a direction with it. For instance, if the area element liesin the x-y plane, it can be considered to be directed along thez–direction.(Conventionally, the direction of the area is taken to be along the outwardnormal.)

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Sθ

In the figure above, the length of the vector~S is chosen to represent the areain some convenient unit and its direction is taken to be alongthe outwardnormal to the area.

We define the flux of the electric field through an area~dS to be given by the scalarproduct

dφ = ~E · ~dS

If θ is the angle between the electric field and the area vector

dφ =| E || dS | cos θ

. For an arbitrary surface S, the flux is obtainted by integrating over all the surfaceelements

φ =

∫

dφ =

∫

S

~E · ~dS

If the electric field is uniform, the angleθ is constant and we have

φ = ES cos θ = E(S cos θ)

Thus the flux is equal to the product of magnitude of the electric field and theprojection of area perpendicular to the field.

3

S

E

θ

Unit of flux is N-m2/C. Flux is positive if the field lines come out of the surfaceand is negative if they go into it.

Solid Angle :The concept of solid angle is a natural extension of a plane angle to three dimen-sions. Consider an area element dS at a distancer from a point P. Letn be theunit vector along the outward normal todS.

The element of the solid angle sub-tended by the area element at P is de-fined as

dΩ =dS⊥

r2

wheredS⊥ is the projection ofd~Salong a direction perpendicular to~r.If α is the angle betweenr and n,then,

dΩ =dS cos α

r2

r

n

^

P dΩ

α

dS

Solid angle is dimensionless. However, for practical reasons it is measured interms of a unit calledsteradian (much like the way a planar angle is measured interms of degrees).The maximum possible value of solid angle is4π, which is the angle subtendedby an area which encloses the point P completely.Example 1:A right circular cone has a semi-vertical angleα. Calculate the solid angle at theapex P of the cone.Solution :The cap on the cone is a part of a sphere of radius R, the slant length of the cone.Using spherical polar coordinates, an area element on the cap is R2 sin θdθdφ,whereθ is the polar angle andφ is the azimuthal angle. Here,φ goes from 0 to2π

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while θ goes from 0 toα.

Thus the area of the cap is

dA = 2πR2

∫ α

0

sin θdθ

= 2πR2(1 − cos α)

Thus the solid angle at P is

dΩ =dA

R2= 2π(1 − cos α)

R

P θ

Exercise :Calculate the solid angle subtended by an octant of a sphere at the centre of thesphere. (Ans. π/2)The flux per unit solid angle is known as theintensity.Example 2An wedge in the shape of a rectangular box is kept on a horizontal floor. Thetwo triangular faces and the rectangular face ABFE are in thevertical plane. Theelectric field is horizontal, has a magnitude8 × 104 N/C and enters the wedgethrough the face ABFE, as shown. Calculate the flux through each of the facesand through the entire surface of the wedge.

0.4 m

0.3m

0.2m

E

A

B

C

DE

F

Solution :The outward normals to the triangular faces AED, BFC, as wellas the normal tothe base are perpendicular to~E. Hence the flux through each of these faces is zero.The vertical rectangular face ABFE has an area 0.06 m2. The outward normal tothis face is perpendicular to the electric field. The flux is entering through this

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face and is negative. Thus flux through ABFE is

φ1 = −0.06 × 8 × 104 = −.48 × 104 N − m2/C

To find the flux through the slanted face, we need the angle thatthe normal to thisface makes with the horizontal electric field. Since the electric field is perpen-dicular to the side ABFE, this angle is equal to the angle between AE and AD,which is cos−1(.3/.5). The area of the slanted face ABCD is 0.1 m2. Thus theflux through ABCD is

φ2 = 0.1 × 8 × 104 × (.3/.5) = +0.48 × 104 N − m2/C

The flux through the entire surface of the wedge isφ1 + φ2 = 0.

Example 3:Calculate the flux through the base of the cone of radiusR.

E

Solution :The flux entering is perpendicular to the base. Since the outward normal to the cir-cular base is in the opposite sense, the flux is negative and isequal to the productof the magnitude of the field and the area of the base, The flux, therefore is,πR2E.

Example 4 :Calculate the flux coming out through the curved surface of the cone in the aboveexample.Solution :

6

dl

h

H r

θ

R

Consider a circular strip of radiusr at a depthh from the apex of the cone. Theangle between the electric field through the strip and the vector ~dS is π−θ, whereθ is the semi-angle of the cone. Ifdl is the length element along the slope, the areaof the strip is2πrdl. Thus,

~E · ~dS = 2πrdl | E | sin θ

We have,l = h/ cos θ, so thatdl = dh/ cos θ. Further,r = h tan θ Substituting,we get

~E · ~dS = 2πh tan2 θ | E | dh

Integrating fromh = 0 to h = H, the height of the cone, the outward flux is| E | H2 tan2 θ = πR2 | E |.Example 5 :A chargeQ is located at the center of a sphere of radiusR. Calculate the fluxgoing out through the surface of the sphere.

dS

By Coulomb’s law, the field due to the chargeQ is radial and is given on thesurface of the sphere by,

~E =1

4πǫ0

Q

R2r

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The direction of the area vector~dS, is also radial at each point of the surface~dS = dSr. The flux

φ =

∫

~E · ~dS

=1

4πǫ0

Q

R2

∫

dS

The integral overdS is equal to the surface area of the sphere, which is,4πR2.Thus the flux out of the surface of the sphere is

φ =1

4πǫ0

Q

R2· 4πR2 =

Q

ǫ0

Note that the flux is independent of the radius of the sphere - acancellationdue to the fact that the surface area of the sphere is proportional to r2 whilethe field is proportional to1/r2. Curiously, the result is valid for any arbitrar-ily shaped surface. Consider a cone of solid angledΩ centered at the charge.This will intersect the arbitrarilyshaped surface in an ellipse whosenormaln makes an angleθ with theoutgoing radial directionr. If thearea of the ellipse isdS,

dΩ =dS cos θ

r2

so that the flux through the cone is

dΦ =1

4πǫ0

q

r2r · ~dS

=1

4πǫ0

q

r2r ·

r2dΩ

cos θn

=1

4πǫ0

qdΩn · r

cos θ

=qdΩ

4πǫ0

(1)

q

dS

rn

^

r

dΩ

Thus the total flux throgh the surface is

Φ =

∫

dΦ =q

4πǫ0

∫

dΩ =q

ǫ0

(2)

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One can generalize this to multiple charges since superposition principle holds forthe electric field. Suppose the total field consists of fieldsE1 due to chargeq1, E2

due toq2 and so on. We have∫

~E · ~dS =

∫

∑

i

~Ei · ~dS =∑

i

qi

ǫ0

One should note that the abovederivation is valid only if the chargesare contained within the volume foronly then the total solid angle be-comes4π. For charges which areoutside the volume, the flux that en-ters the volume also leaves it andthough the areas which the cone in-tersects are different, the solid anglesare the same. This leads to cancella-tion of the flux and the contributionto the flux from a charge which re-sides outside the volume is zero.

r

dΩq

r

n

n dS

dS1

1 2

2

GAUSS’S LAW - Integral form

The flux calculation done in Example 4 above is a general result for flux out ofany closed surface, known as Gauss’s law.Total outward electric flux φ through a closed surfaceS is equal to1/ǫ0 timesthe charge enclosed by the volume defined by the surfaceS

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E

E

E

dS

Mathematicaly, the surface integral of the electric field over any closed surface isequal to the net charge enclosed divided byǫ0

∮

~E · ~dS =Qenclosed

ǫ0

(3)

• The law is valid for arbitry shaped surface, real or imaginary.

• Its physical content is the same as that of Coulomb’s law.

• In practice, it allows evaluation of electric field in many practical situa-tions by forming imagined surfaces which exploit symmetry of the problem.Such surfaces are calledGaussian surfaces.

GAUSS’S LAW - Differential form

The integral form of Gauss’s law can be converted to a differential form byusing the divergence theorem. IfV is the volume enclosed by the surface S,

∮

S

~E · ~dS =

∫

V

~∇ · ~Edv (A)

If ρ is the volume charge density,

Q =

∫

V

ρdv (B)

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Thus we have∇ · ~E =

ρ

ǫ0

(4)

Direct Calculation of divergence from Coulomb’s Law :We will use the field expression

~E(~r) =1

4πǫ0

∫

(~r − ~r′)

| ~r − ~r′ |3ρ(~r′)d3r

to directly evaluate the divergence of the electric field. Since the differentiationis with respect to~r while the integration is with respect to~r′, we can take thedivergence inside the integral,

~∇ · ~E(~r) =1

4πǫ0

∫

∇r ·(~r − ~r′)

| ~r − ~r′ |3ρ(~r′)d3r

A simple minded calculation of divergence is as follows :

∇r ·(~r − ~r′)

| ~r − ~r′ |3= (∇r · ~r)

1

| ~r − ~r′ |3+ (~r − ~r′) ·

(

∇r

1

| ~r − ~r′ |3

)

Divergence of~r is equal to 3

∇r · ~r =∂

∂xx +

∂

∂yy +

∂

∂zz = 3

∇r

1

| ~r − ~r′ |3= −3

~r − ~r′

| ~r − ~r′ |5

Thus it would seem that

∇r ·(~r − ~r′)

| ~r − ~r′ |3= 0

i.e. ∇ · ~E = 0, which violates Gauss’s Law, that∇ · ~E = ρ/ǫ0. The problemarises because the function1/ | ~r − ~r′ | has a singularity at~r = ~r′. This point hasto be taken care of with care. Except at this point the divergence of the integral isindeed zero.

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Thus we can shrink the range of in-tegral till it becomes a small spherearound the point~r(x, y, z). Sinceρ(x′, y′, z′) is continuous, we mayreplace the density on the surface ofthe small sphere by the value of den-sity at the centre, so that the densityterm can come out of the integral,leaving,

~∇· ~E(~r) =1

4πǫ0

ρ(~r)

∫

∇r·(~r − ~r′)

| ~r − ~r′ |3d3r

r − r’

(x,y,z)

(x’,y’,z’)

Since the divergence with respect to~r is being taken of a function which onlydepends on the difference~r − ~r′, we may replace∇r → −∇′

r, which gives

~∇ · ~E(~r) = −1

4πǫ0

ρ(~r)

∫

∇r′ ·(~r − ~r′)

| ~r − ~r′ |3d3r

The volume integral on the right may be converted to a surfaceintegral using thedivergence theore,

~∇ · ~E(~r) = −1

4πǫ0

ρ(~r)

∫

(~r − ~r′)

| ~r − ~r′ |3· d~S

where the integral is over the surface of the sphere. If the radius of the sphere betaken asr0, ~r − ~r′ = −r0n and ~dS = dSn, Hence

~∇ · ~E = +1

4πǫ0

ρ(~r)

∫

dS

r2

0

=1

4πǫ0

ρ(~r) · 4πr2

0=

ρ(~r)

ǫ0

Example 1: A sphere of radiusR contains a continuous charge distributionρ(r).The electric field at a distancer from the centre of the sphere is~E = kr3r.(a) Find the charge density.Soln. :

~∇ ~E =1

r2

∂

∂r(r2Er) =

1

r2

∂

∂r(kr5) = 5kr2 = ρ/ǫo

Thusρ = 5kr2ǫo. (b) Find the total charge contained in the sphere.Soln. :The total charge is obtained by integrating the charge density

∫ R

0

ρdτ = 5ǫok

∫

r24πr2dr = 4πkǫ0R5

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The same result is also obtained by the surface integral of the electric field :∫

~E · ~dS =

∫

kR3r · (R2 sin θdθdφr) = 4πkR5 =Q

ǫ0

The two results are consistent.Example 2 :

Calculate the flux through the shadedarea (face of a cube of sidea) when achargeq is located at one of the dis-tant corners from the side.

q

Solution :If the charge were located at the centre of the cube instead ofthe corner, the fluxwould have beenq/6ǫ0, by symmetry. To use this symmetry consider the givencube to be a part of a bigger cube of side2a × 2a, as shown, so that the chargeqis in the centre of the bigger cube.

The flux through each face of thebegger cube is nowq/6ǫ0. Be-cause the side of the bigger cubeconsists of four identical faces, theflux through one fourth of the face isclearlyq/24ǫ0.

Applications of Gauss’s Law

Field due to a uniformly charged sphere of radiusR with a chargeQ

By symmetry, the field is radial. Gaussian surface is a concentric sphere of radiusr. The outward normals to the Gaussian surface is parallel to the field ~E at everypoint. Hence

∫

~E · ~dS = 4πr2 | E |

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Q

r

R

E

R

rQ

E

r >R r < R

For r > R,

4πr2 | E |=Q

r2

so that~E =

Q

4πr2r

The field outside the sphere is what it would be if all the charge is concentrated atthe origin of the sphere.

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For r < R, a fractionr3/R3 of thetotal charge is enclosed within thegaussian surface, so that

4πr2 | E |=1

ǫ0

Qr3

R3

The field inside is

~E =Q

4πǫ0

r

R3r

E in a

rbitra

ry u

nits

r--->

Field due to sphere with charge density ρ

R

Exercise :Find the electric field both inside and outside a spherical shell of radiusR carryinga uniform chargeQ.

Example 3 :Find the electric field inside a sphere of radiusR which carries a charge densityρ = kr wherer is the distance from the origin andk is a constant.Solution :By symmetry the field is radial. Take the gaussian surface to be a sphere of radiusR. The flus is4πr2 | E |.

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r

R

The charge enclosed by the gaussian surface is

Q =

∫ r

0

ρ(r)d3r =

∫ r

0

ρ(r)4πr2dr

= 4πk

∫ r

0

r3dr

= πkr4

Thus~E =

1

ǫ0

kr2

4r

(what is the dimension ofk ?)

Exercise : A hollow spherical shell carries a charge densityρ = k/r2 fora ≤ r ≤ b. Calculate the field at all points. (Ans. Forr < a, field is zero, fora < r < b, | E |= k(r − a)/ǫ0r

2, and forr > b, | E |= k(b − a)ǫ0r2.)

Field due to an infinite line charge of linear charge densityλ

Gaussian surface is a cylinder of radiusr and lengthL.By symmetry, the field has the same magnitude at every point onthe curved sur-face and is directed outwards. At the end caps,~E is perpendicular to~dS every-where and the flux is zero. For the curved surface,~E and ~dS are parallel,

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dS1

dS2E

L

++++++++++++++++

∮

~E · ~dS = | E | .2πrL

=Q

ǫ0

=λL

ǫ0

Thus~E =

λ

2πǫ0rρ

whereρ is a unit vector perpendicular to the line,directed outwardfor positive linecharge and inward for negative line charge.

Exercise :Find the electric field both inside and outside a long cylinder of radiusR carryinga uniform volume charge densityρ.(Hint : Take the gaussian surface to be a finite concentric cylinder of radiusr (withr < R andr > R), as shown)

+ + + + + + +++ +

3 Field due to an infinite charged sheet with surface charge density σ

Choose a cylindricalGaussian pillboxof heighth (with h/2 above the sheet andh/2 below the sheet) and radiusr.

17

dS

E

+

+ +

+ + +

+ + +

+

+

+

+ + +

+

+

+

+ +

+

+ + + +

+ L

σ

dSE

E

r

dS

The amount of charge enclosed is area times the surface charge density, i.e.,Q =πr2σ. By symmetry, the field is directed perpendicular to the sheet, upward atpoints above the sheet and downward for points below. There is no contributionto the flux from the curved surface. The flux from the two end faces isπr2 | E |each, i.e. a total outward flux of2πr2 | E |. Hence

2πr2 | E |=Q

ǫ0

=πr2σ

ǫ0

so that~E =

σ

2ǫ0

n

where n is a unit vector perpendicular to the sheet, directed upwardfor pointsabove and downwards for points below (opposite, if the charge density is nega-tive).

Exercise :Find the electric field in the region between two infinite parallel planes carryingcharge densities+σ and−σ.

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+

+

+

+

+

++

+

+σ −σ

Exercise :A very long cylinder carries a charge densityρ = kr, wherer is the distance fromthe axis of the cylinder. Find the electric field at a distancer < R. (Ans.(1/3ǫ0)kr2r)

Example 4 :Two spheres of radiusR each overlap such that the distance between their centresis separated by a distance2R − s. Show that the field in the overlapping region isconstant.Solution :

O O’

s

θφ

P

Q

The figure shows the field at a point P in the overlap region due to the two spheres.Taking the expression for field at a point inside the sphere and resolving into x-and y- components, The x- components reinforce while the y-components are in

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the opposite directions. It can be seen,

Ey =Q

4πǫ0

1

R3(OP sin θ − O′P sin φ)

Using the property of triangles (the ratio of the sides is equal to the ratio of thesine of opposite anglesa/ sin A = b/ sin B) the y-component is seen to vanish.The x-component of the field is

Ex =Q

4πǫ0

1

R3(OP cos θ + O′P cos φ)

=Q

4πǫ0

1

R3(OQ + O′Q)

=Q

4πǫ0

1

R3s

Thus the field depends only on the distance between the centres and is constant.

Example 5 :A sphere of radius R has a cavity of radius a inside it. The sphere has uniformcharge densityρ spread over its volume. Show that the field inside the cavity isconstant.Solution :

One can use superposition princi-ple to solve this problem. Fill upthe cavity with equal and oppositecharge distribution. The problenmthen is equivalent to the field due toa sphere of radiusR and charge den-sity ρ and a smaller sphere of radiusa, but with a charge densiuty−ρ.

R

a

dr

O

PO’

We calculate the fields due to these two spheres at~r (with respect to O, the centre

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of the larger sphere, The field due to larger sphere

~E1 =Q

4πǫ0

~r

R3

=ρ

4πǫ0

4π

3R2

~r

R3

=ρ

3ǫ0

~r

By identical argument, the field due to smaller sphere (the point P is at~r − ~d withrespect to the centre of the smaller sphere),

~E2 = −ρ

3ǫ0

(~r − ~d)

Adding,~E =

ρ

3ǫ0

~d = constant

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