Electrostatics Concept Page - 8

Law

Gauss's Law

Gauss law relates the net flux ϕ$\varphi$ of an electric field through a closed surface (a Gaussian surface) to the net charge qenc$q_{enc}$ that is enclosed by that surface.
ϵoϕ=qenclosed${ϵ}_o\varphi =q_{enclosed}$
By the definition of flux, we can also write Gauss law as
∮E⃗ .dS⃗ =qenclosedε0$\oint \overrightarrow{E}.d\overrightarrow{S}={\displaystyle \frac{q_{enclosed}}{{\epsilon }_0}}$
where qenclosed$q_{enclosed}$ is the net charge enclosed by the surface through which flux is to be found.

Law

Derivation from Coulomb's Law: Field due to an internal charge

E=q4πϵ0r2$E={\displaystyle \frac{q}{4\pi {ϵ}_0{r}^2}}$
where q is a charge placed at a point inside a closed surface 
r is the distance between q and the point on the surface where the electric field is calculated.
Δϕ=q4πϵ0r2ΔScosα=q4πϵ0ΔΩ$\mathrm{\Delta }\varphi ={\displaystyle \frac{q}{4\pi {ϵ}_0{r}^2}}\mathrm{\Delta }Scos\alpha ={\displaystyle \frac{q}{4\pi {ϵ}_0}}\mathrm{\Delta }\mathrm{\Omega }$
where α$\alpha$ is the angle between the outward normal vector and the electric field at the point on the surface and ΔΩ$\mathrm{\Delta }\mathrm{\Omega }$ is the solid angle subtended at q by the area element.
Then total flux ϕ=q4πϵ04π=qϵ0$\varphi ={\displaystyle \frac{q}{4\pi {ϵ}_0}}4\pi ={\displaystyle \frac{q}{{ϵ}_0}}$
Flux due to an external chargeHere ∑ΔΩ=0$\sum \mathrm{\Delta }\mathrm{\Omega }=0$
ϕ=0$\varphi =0$
Field due to a combination of charges
Consider a system of charges q1,q2,q3,....qn$q_1,q_2,q_3,....q_n$ inside a closed surface and charges Q1,Q2,Q3,.....,Qn$Q_1,Q_2,Q_3,.....,Q_n$ placed outside the surface. 
Then E⃗ =E⃗ 1+E⃗ 2+E⃗ 3+.....+E⃗ n+E⃗ ′1+E⃗ ′2+E⃗ ′3+....+E⃗ ′n$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3+.....+{\overrightarrow{E}}_n+{\overrightarrow{E}}_1^{{}^{\prime }}+{\overrightarrow{E}}_2^{{}^{\prime }}+{\overrightarrow{E}}_3^{{}^{\prime }}+....+{\overrightarrow{E}}_n^{{}^{\prime }}$
where E⃗ i${\overrightarrow{E}}_i$ and E⃗ ′i${\overrightarrow{E}}_i^{{}^{\prime }}$ are the electric fields due to qi$q_i$ and Qi$Q_i$  respectively.
Φ=∮E⃗ .dS⃗ =∮E1−→.dS⃗ +∮E2−→.dS⃗ +∮E3−→.dS⃗ +.....+∮En−→.dS⃗ +∮E′1−→+∮E′2−→+∮E′3−→+...+∮E′n−→$\mathrm{\Phi }=\oint \overrightarrow{E}.d\overrightarrow{S}=\oint \overrightarrow{E_1}.d\overrightarrow{S}+\oint \overrightarrow{E_2}.d\overrightarrow{S}+\oint \overrightarrow{E_3}.d\overrightarrow{S}+.....+\oint \overrightarrow{E_n}.d\overrightarrow{S}+\oint \overrightarrow{E_1^{{}^{\prime }}}+\oint \overrightarrow{E_2^{{}^{\prime }}}+\oint \overrightarrow{E_3^{{}^{\prime }}}+...+\oint \overrightarrow{E_n^{{}^{\prime }}}$
Φ=q1ϵ0+q2ϵ0+q3ϵ0+........+qnϵ0+0+.....+0$\mathrm{\Phi }={\displaystyle \frac{q_1}{{ϵ}_0}}+{\displaystyle \frac{q_2}{{ϵ}_0}}+{\displaystyle \frac{q_3}{{ϵ}_0}}+........+{\displaystyle \frac{q_n}{{ϵ}_0}}+0+.....+0$
Φ=1ϵ0∑qi$\mathrm{\Phi }={\displaystyle \frac{1}{{ϵ}_0}}\sum q_i$
∮E⃗ .dS⃗ =qinϵ0$\oint \overrightarrow{E}.d\overrightarrow{S}={\displaystyle \frac{q_{in}}{{ϵ}_0}}$

Example

Application of Gauss law

Fig, which shows two particles, with charges equal in magnitude but opposite in sign, and the field lines describing the electric fields the particles set up in the surrounding space. Four Gaussian surfaces are also shown, in cross section. Let us consider each in turn
Surface 1 - The electric field is outward for all points on this surface. Thus, the
flux of the electric field through this surface is positive, and so is the net charge within the surface, as Gauss law requires.
Surface 2 - The electric field is inward for all points on this surface.Thus, the flux
of the electric field through this surface is negative and so is the enclosed charge, as Gauss law requires.
Surface 3 - This surface encloses no charge, and thusqenclosed=0$q_{enclosed}=0$. Gauss law requires that the net flux of the electric field through this surface be zero. That is reasonable because all the field lines pass entirely through the surface, entering it at the top and leaving at the bottom.
Surface 4 - This surface encloses no net charge, because the enclosed positive
and negative charges have equal magnitudes. Gauss law requires that the net
flux of the electric field through this surface be zero. That is reasonable because there are as many field lines leaving surface S4$S_4$ as entering it.

What would happen if we were to bring an enormous charge Q up close to surface S4$S_4$ in Fig.? The pattern of the field lines would certainly change, but the net flux for each of the four Gaussian surfaces would not change. Thus, the value of Q would not enter Gauss law in any way, because Q lies outside all four of the Gaussian surfaces that we are considering.

Law

Gaussian surface

The surface that we choose for the application of Gauss's law is called the Gaussian surface.The Gaussian surface doesn't pass through any discrete charge. This is because electric field due to a system of discrete charges is not well defined at the location of any charge.However, the Gaussian surface can pass through a continuous charge distribution.

Example

Electric flux through a solid angle

Example:
A point charge Q is kept at the center of a sphere of radius 1 m. Find the flux passing through a surface of area 1m2$1m^2$ on the sphere.
Solution:
Solid angle subtended by the surface area 1m2$1m^2$ at the center is Ω=14π×12×4π=1steradian$\mathrm{\Omega }={\displaystyle \frac{1}{4\pi \times 1^2}}\times 4\pi =1steradian$
Flux through surface area of 1m2$1m^2$ is ϕ=ϕtotal×Ω4π$\varphi ={\varphi }_{total}\times {\displaystyle \frac{\mathrm{\Omega }}{4\pi }}$
ϕ=Q4πεo$\varphi ={\displaystyle \frac{Q}{4\pi {\epsilon }_o}}$

Result

Electric field analogy of charged sphere (outside) and a point charge

Electric field due to a charged sphere outside it is given by, 
E=Q4πεor2$E={\displaystyle \frac{Q}{4\pi {\epsilon }_o{r}^2}}$
This expression is exactly same as the electric field of a point charge.
Hence, a charged sphere behaves as a point charge for a point outside it.
Note:
The above statement is true only for point charge and charged sphere and is invalid for other charge distributions.

Formula

Electric field intensity due to a charged non-conducting sphere

E=q4πϵ0r2$E={\displaystyle \frac{q}{4\pi {ϵ}_0{r}^2}}$              (r≥R)$(r\ge R)$
E=qr4πϵ0R3$E={\displaystyle \frac{qr}{4\pi {ϵ}_0{R}^3}}$            (r<R)$(r<R)$

Diagram

Electric field intensity vs distance for a charged non-conducting sphere

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8  Page 9  Page 10
Page 11  Page 12  Page 13  Page 14  Page 15