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Gauss’s Law of Electrostatics

Statement : “The total normal electric flux over a closed surface in an electric field is equal to 1/ƹ0 times the total charge enclosed by that surface.”

Mathematically it may be expressed as

Proof. (l) When the charge lies inside the closed surface: Let us consider a source producing the field is a point charge + q situated at 0 inside the closed surface as shown in Figure 7.17.

 

FIGURE 7.17

 

Let dS be an infinitesimal element of the surface at point P and OP = r. As shown in Figure 7.17, the electric field strength vector E makes angle 8 with the unit vector n drawn normal to the surface element dS surrounding point P. The surface integral of the normal component of this electric field over the closed surface is given by ʃ ʃs E. n dS.

 

The electric field strength E at the point P is given by

Now, it can be seen that the quantity ( r .n/r  dS) gives the projection of area dS on theplane perpendicular to r and therefore.

 

projected area  r2 =r. n dS  /r3= dw

 

where dw is the solid angle subtended by dS at 0.

 

From equation (1) and (2) we get

ʃʃS E .n ds =1/4πƹ0  q  ʃ ʃsdw

 

But ʃ ʃSdw = 4π = solid angle subtended by entire closed surface at an internal point

ʃʃS E .n ds =1/4πƹ0 q .4 =q/ƹ 0

 

 

This result is known as Gauss’s law for a single point charge enclosed by the surface.

If several point charges q1, q2, q3 . . . . qn be enclosed by the surface 5, then total electric field is given by L qi (= q1 + q2 + q3 + . …. .. + qn). Each charge subtends a full solid angle 4n and equation (4) becomes ·

  1. (ii) When the charge is situated outside the closed surface: If the charge q is outside the surface (Figure 7.18), then the surface 5 can be divided into areas S1, S2, S3 and S4each of which subtends the same solid angle at the charge q. But at S1 and S3 the directions of the  outward drawn normal are away from q, while at S2 and S4 they are toward q. Therefore the contributions of two pairs (S1, S3) and (S2, S4) to the surface integral are equal and opposite. Asa result the Net surface integral of the normal component of the electric field Evanishes, i.e.,

ʃʃS E .n ds =ʃ ʃS E .ds = 0

 

equations (4) and (6) represents the integral form of the Gauss’s law.

7.12.1 Differential Form of Gauss’s law

 

Let a charge q be distributed over a volume V of the closed surface 5 and p be the chargedensity; then the charge q may be given as

 

q = ʃ ʃʃvpdV = ʃ vpdV

 

Thus the total flux through the surface S

According to Gauss divergence theorem the surface integral may be converted into volume integral as

Hence  From equation (2) ,we obtain

ʃv div E dv =1/ƹ0 ʃ v p dv

div E =1/ƹ 0 P

∆ .E =1/ƹ0 P

 

As the displacement vector D is defined as

D = ƹ0 E, we have

∆.D = div D = p

Equations (4) and (5) are differential form of Gauss’s law of electrostatics

7 .12.2 Applications of Gauss’s law of Electrostatics

1.      Derivation of Coulomb’s law of electrostatics from Gauss’s law: Consider twopoint charges q1 and q2 separated by a distance ‘r’. Draw a Gaussian surface ofsphere of radius ‘r’ with q1 as centre. E must be normal tothis surface and must have same magnitude for all pointson the surface. E and dS at any point on the drawnGaussian surface are directed radially outward. The angle , between them is zero.

 

E .ds E ds cos θ =E ds cos 00=E ds

Hence ,by gauss’s theorem

E .4πr2 q1 /ƹ0 or E =q1 /4πƹ0r2

 

The force on the charge q2 = F = Eq2 = q1q2/4πƹ0r2, which is the Coulomb’s law.

1.      Field around a charged straight conductor (Line Charge): Consider a linearcharged conductor. Let the linear charge density be λ Imagine a Gaussian cylinderof radius ‘y’ and length ‘I’,closed at each end by plane caps normal to the axis. Bysymmetry, all the lines of force go radially outward.

 
As shown in Figure 7.20, no lines cross the flat ends of thecylinder and hence the ʃ E dS over these end surfaces are zero.

Over the remaining surface of cylinder, the field is uniform and outward.

 

. . Applying Gauss theorem

ϕ=q/ε0 =lλ/ε0

Net flux passing over the Gaussian surface

From equations (1) and (2),= E (2ny) l or

But                  q /ε0 =σ (2πal)/ε0=(λ/2πa) (2πa)/ε0 =λl

From equations (1) and (2),

Λl /ε0 = E (2πε0r)

4. Field due to a hollow spherical conductor: Let us consider a hollow sphericalconductor of radius R with surface charge density cr. Let q be the net charge of thesurface.

 

q = 4 π r2σ.

 

(i) Point outside the sphere: Consider a point P situated outside the sphere at adistance ‘r’. Imagine a Gaussian spherical surface talking ‘r’ as radius.

 

Area of the Gaussian sphere = 4πr2

By Gauss theorem

Flux over the Gaussian sphere = E ʃdS

E .4πr2 =q /ε0

E =q /4πε0r2

 

(ii) Point on the sphere: When point Pis on the sphere r = R then,

 

E =q /4πε0r2

(iii) Point inside the sphere: For any point inside the sphere, the electric field is zero, since there is no charge inside.

If we draw a Gaussian spherical surface of radius ‘r’,the distance between the centre of the sphere and the pointP as radius, the sphere contains no charge. Hence the fieldat any point inside a charged hollow sphere is zero

 

5. Field due to a solid non-conducting sphere: Let us consider a solid sphericalconductor of radius ‘R’ with volume density ‘p’. The net charge in the sphere,

 

q = 4/3 πR3 p.

 

(i) Point outside the sphere: Consider a point P at adistance ‘r’ from the centre of the sphere. Imagine,a Gaussian spherical surface with ‘r’ as radius.

 

Applying Gauss theorem

E (4πr2) =q/ε0

E =q /4πε0r2

 

(iii) Point inside the sphere: To find the fieldintensity at P inside the solid sphere, considera Gaussian spherical surface with radius ‘r1 ‘.

 

The solid sphere consists of two parts

(i) a solid sphere of radius ‘r1

(ii) a shell of thickness ‘(R – r1)’.

 

The point P is just on the surface of solid sphere ofradius ‘r1‘.

The field due to the solid sphere is obtained as

E (4πr12) =q10

Where              q1=4/3 π r13 p

Then                E (4πr12) = (4/3 π r13 p) /ε0

E =pr1 /3ε0

Q=4/3 π R3 p,

P=q /4/3 πR3

The contribution of the outer shell towards the field is zero, since there is no fieldinside a charged hollow sphere. Hence the field intensity at a point inside the charged solidsphere is given by

 

E=q r1/4πε0R3 =Pr1/3ε0

 

i.e., the field is directly proportional to the distance of the point from the centre of the

charged sphere.

 

6. Electric field due to a charged conductor (Coulomb’s theorem)

 

Statement : The electric field at any point near a charged conductor is 1/c.0 times the surface density of charge on the surface.

 

Proof : Let ‘AB’ be a large conducting surface and σ be the surface charge density. To find the field at a point P,infinitely close to the charged surface, imagine a small cylinder of cross-section ‘S’ drawn with its faces parallel to the charged surface. Le one of the faces pass through P and the other through Q inside the conductor as shown in Figure 7.26. Normal component of E, through the sides of the cylinder, formed by faces enclosing P and Q i.e., Gaussian surface is zero.

Since the charge resides on the surface of the conductor, there is no charge on the face enclosing ‘Q’ and therefore normal component of E through the area dS is also zero. Hence the normal component of ‘E’ is only through containing ‘P’.

 

or

 

According to Gauss theorem,

Eds =σds/ε0

E = σ/ε0