**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 ʃ_{s} 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 r^{2} =r. n dS /r_{3}= dw

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

From equation (1) and (2) we get

ʃʃ_{S} E .n ds =1/4πƹ_{0} q ʃ ʃ_{s}dw

But ʃ ʃ_{S}dw = 4π = solid angle subtended by entire closed surface at an internal point

ʃʃ_{S} E .n ds =1/4πƹ_{0} 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 q_{1}, q_{2}, q_{3} . . . . q_{n} be enclosed by the surface 5, then total electric field is given by L qi (= q_{1} + q_{2} + q_{3} + . …. .. + qn). Each charge subtends a full solid angle 4n and equation (4) becomes ·

**(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 S_{1}, S_{2}, S_{3}and S_{4}each of which subtends the same solid angle at the charge q. But at S_{1}and S_{3}the directions of the outward drawn normal are away from q, while at S_{2}and S_{4}they are toward q. Therefore the contributions of two pairs (S_{1}, S_{3}) and (S_{2}, S_{4}) 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.,

ʃʃ_{S} E .n ds =ʃ ʃ_{S} 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 ʃʃ_{v}pdV = ʃ _{v}pdV

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

ʃ_{v} div E dv =1/ƹ_{0} ʃ _{v} p dv

div E =1/ƹ _{0} P

∆ .E =1/ƹ_{0} P

As the displacement vector D is defined as

D = ƹ_{0} 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 q_{1} and q_{2} separated by a distance ‘r’. Draw a Gaussian surface ofsphere of radius ‘r’ with q_{1} 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 0^{0}=E ds

Hence ,by gauss’s theorem

E .4πr^{2} q1 /ƹ^{0} or E =q1 /4πƹ^{0}r^{2}

The force on the charge q_{2} = F = Eq_{2} = q_{1}q_{2}/4πƹ_{0}r^{2}, 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/ε_{0} =lλ/ε_{0}

Net flux passing over the Gaussian surface

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

But q /ε0 =σ (2πal)/ε_{0}=(λ/2πa) (2πa)/ε_{0} =λl

From equations (1) and (2),

Λl /ε_{0 = }E (2πε_{0}r)

**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 r^{2}σ.

(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 = 4πr^{2}

By Gauss theorem

Flux over the Gaussian sphere = E ʃdS

E .4πr^{2} =q /ε_{0}

E =q /4πε_{0}r^{2}

^{ }

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

E =q /4πε_{0}r^{2}

(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 πR^{3} 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 (4πr^{2}) =q/ε_{0}

E =q /4πε_{0}r^{2}

^{ }

(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 ‘r_{1}‘

(ii) a shell of thickness ‘(R – r_{1})’.

The point P is just on the surface of solid sphere ofradius ‘r_{1}‘.

The field due to the solid sphere is obtained as

E (4πr_{1}^{2}) =q_{1} /ε_{0}

Where q_{1}=4/3 π r_{1}^{3} p

Then E (4πr_{1}^{2}) = (4/3 π r_{1}^{3} p) /ε_{0}

E =pr_{1} /3ε_{0}

Q=4/3 π R^{3} p,

P=q /4/3 πR^{3}

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 =Pr1/3ε_{0}

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/ε_{0}

E = σ/ε_{0}