Electrostatics Concept Page - 9

Example

Electric field due to non-uniformly charged sphere

Example:
A non-uniformly charged sphere of radius R$R$ has a charge density ρ=ρo(r/R)$\rho ={\rho }_o(r/R)$ where ρo${\rho }_o$ is constant and r$r$ is the distance from the center of the sphere. Find the electric field inside and outside the sphere.
Solution:
Electric field lines will be directed radially outwards as the charge is distributed radially uniformly.
For 0<r<R$0<r<R$,
      Charge enclosed in an infinitesimally small spherical shell of thickness dr$dr$ at a distance r$r$ is:
      dQenclosed=ρ(r)dV$d{Q}_{enclosed}=\rho (r)dV$
      dQenclosed=ρorR4πr2dr$d{Q}_{enclosed}={\rho }_o{\displaystyle \frac{r}{R}}4\pi r^{2}dr$
      Charge enclosed upto distance r$r$ is:
      Q=∫dQenclosed=∫r0ρo4πr3Rdr${\displaystyle Q=\int d{Q}_{enclosed}={\int }_0^r{\rho }_o{\displaystyle \frac{4\pi r^3}{R}}dr}$
      Q=ρoπr4R$Q={\displaystyle \frac{{\rho }_o\pi r^4}{R}}$                 
      Applying gauss law at a distance r$r$, 
      E×4πr2=ρoπr4R$E\times 4\pi r^2={\displaystyle \frac{{\rho }_o\pi r^4}{R}}$
      E=ρor24R$E={\displaystyle \frac{{\rho }_o{r}^2}{4R}}$
For r>R$r>R$,
      Charge enclosed is: 
      Q=ρoπR3$Q={\rho }_o\pi R^3$
      Electric field outside is hence given by:
      E=ρoR34r2$E={\displaystyle \frac{{\rho }_o{R}^3}{4r^2}}$

Definition

Electric field near and outside a charged conductor

For a conductor, all charge is distributed on the outer surface of the conductor such that the surface charge density is more at the regions with smaller radius of curvature (example pointed regions) and all points are at same potential. When the point is close to the surface, electric field is given by:
E=σεo$E={\displaystyle \frac{\sigma }{{\epsilon }_o}}$
where σ$\sigma$ is the surface charge density.

Example

Application of Gauss Law in a given non uniform electric field

Example:
An electric field given by E⃗ =4i^−3(y2+2)j^$\overrightarrow{E}=4\hat{i}-3(y^2+2)\hat{j}$ passes Gaussian cube of side 1m placed with one corner is at origin such that its sides represents x, y and z axes. Find the magnitude of net charge enclosed within the cube.Solution:
The flux passing through the surfaces parallel to x-y plane is zero as their is no z component of E.
The flux passing through the surfaces parallel to y-z plane is zero as the x component of E is a constant. Flux going in goes out, so no net flux.
The flux passing through the surface parallel to x-z plane is given by,
ϕ1=−6${\varphi }_1=-6$  and ϕ2=−9${\varphi }_2=-9$
ϕnet=−9+6=−3${\varphi }_{net}=-9+6=-3$
Magnitude of charge enclosed =|ϕnet|×ϵ0=3ϵ0$=|{\varphi }_{net}|\times {ϵ}_0=3{ϵ}_0$

Definition

Total normal electric induction

The total number of lines of induction leaving a surface normally is called total normal induction. Hence, 
ϕ=∫E⃗ .d⃗ S$\varphi =\int \overrightarrow{E}.\overrightarrow{d}S$

Example

Electric flux through a closed surface enclosing a point charge

Φ=qϵ0$\mathrm{\Phi }={\displaystyle \frac{q}{{ϵ}_0}}$
where q is the charge enclosed inside a closed surface
Determine the electric flux for a gaussian surface that contains 100 millions electrons?
By using formula of electric flux
Φ=qϵ0$\mathrm{\Phi }={\displaystyle \frac{q}{{ϵ}_0}}$
Where q = charge on electron *no. of electrons = 1.6×10−11$1.6\times {10}^{-11}$
Φ=1.6×10−118.85×10−12$\mathrm{\Phi }={\displaystyle \frac{1.6\times {10}^{-11}}{8.85\times {10}^{-12}}}$
∴Φ=1.8Nm2C$\therefore \mathrm{\Phi }=1.8{\displaystyle \frac{N{m}^2}{C}}$

Example

Application of Gauss Law in cubical surface

Example:
At the center of a cubical box + Q charge is placed. Find the value of total flux that is coming out of each face.
Solution:
By symmetry, the flux through each of the surface is same.
Total flux, ϕ=Qεo$\varphi ={\displaystyle \frac{Q}{{\epsilon }_o}}$
Flux through each surface, ϕs=Q6εo${\varphi }_s={\displaystyle \frac{Q}{6{\epsilon }_o}}$

Example

Finding electric flux through an area by completing a symmetrical gaussian surface

Example:
A charge Q is placed at the mouth of a conical flask. Find the flux of the electric field through the flask.
Solution:A new conical flask can be assumed as shown in the attached figure. 
Total flux, ϕ=Qεo$\varphi ={\displaystyle \frac{Q}{{\epsilon }_o}}$
Using symmetry, flux through each conical surface is same. Hence, flux through the upper conical surface =ϕ2=Q2εo$={\displaystyle \frac{\varphi }{2}}={\displaystyle \frac{Q}{2{\epsilon }_o}}$

Example

Electric potential due to a spherically symmetric distribution of charge

Example:
Consider a spherical shell of radius R$R$ with a charge of Q$Q$. Find electric potential inside and outside the spherical shell.
Solution:
For r>R$r>R$, 
      V=Q4πεor$V={\displaystyle \frac{Q}{4\pi {\epsilon }_{o}r}}$
      In this region, spherical shell acts similar to point charge.
For r≤R$r\le R$,
       Vr=VR=Q4πεoR$V_r=V_R={\displaystyle \frac{Q}{4\pi {\epsilon }_{o}R}}$
       In this region, there is zero electric field and hence electric potential is constant.
  

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