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Electric Charges and Fields — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields5 Marks
Question
Answer the following questions:
  1. Define electric flux. Write its $SI $ unit.
  2. Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
  3. How is the field directed if:
  • The sheet is positively charged.
  • Negatively charged?
✓

Answer

  1. Electric flux: It is defined as the total number of electric field lines passing through an area normal to its surface.
$\Phi=\oint\vec{\text{E}}.\vec{\text{dS}}$
The $SI$ unit is $Nm^2/ C$ or volt $-$ metre.
  1. Image
Let electric charge be uniformly distributed over the surface of a thin, non-conducting infinite sheet.
Let the surface charge density $($i.e., charge per unit surface area$)$ be $\sigma$ .
​​​​​​​We need to calculate the electric field strength at any point distant r from the sheet of charge.
To calculate the electric field strength near the sheet, we now consider a cylindrical Gaussian surface bounded by two plane faces $A$ and $B$ lying on the opposite sides and parallel to the charged sheet and the cylindrical surface perpendicular to the sheet$ ($fig$)$. By symmetry the electric field strength at every point on the flat surface is the same and its direction is normal outwards at the points on the two plane surfaces and parallel to the curved surface.
Total electric flux,
or $\oint\limits_\text{S}\vec{\text{E}}.\vec{\text{dS}}=\oint\limits_{\text{S}_1}\vec{\text{E}}.\vec{\text{dS}_1}+\oint\limits_{\text{S}_2}\vec{\text{E}}.\vec{\text{dS}_2}+\oint\limits_{\text{S}_3}\vec{\text{E}}.\vec{\text{dS}_3}$
$\oint\limits_\text{S}\vec{\text{E}}.\vec{\text{dS}}=\oint\limits_{\text{S}_1}\text{E dS}_1\cos0^\circ+\oint\limits_{\text{S}_2}\text{E dS}_2\cos0^\circ+\oint\limits_{\text{S}_3}\text{E dS}_3\cos90^\circ$
$=\text{E}\oint\text{dS}_1+\text{E}\oint\text{dS}_2=\text{Ea}+\text{Ea}=2\text{Ea}$
$\therefore$ Total electric flux $= 2Ea$
As $\sigma$ is charge per unit area of sheet and a is the intersecting area, the charge enclosed by Gaussian surface $=\sigma\text{a}$
Total electric flux $=\frac{1}{\epsilon_0}\times  \ ($total charge enclosed by the surface$)$
i.e., $2\text{Ea}=\frac{1}{\epsilon_0}(\sigma\text{a})$
$\therefore\ \text{E}=\frac{\text{p}}{2\epsilon_0}.$
Thus electric field strength due to an infinite flat sheet of charge is independent of the distance of the point.
  1.  
  • If $\sigma$ is positive, $\vec{\text{E}}$ points normally outwards/ away from the sheet.
  • If $\sigma$ is negative, $\vec{\text{E}}$ points normally inwards/ towards the sheet.

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