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$$\begin{gathered} \mathrm{Charge} \mathrm{on} \mathrm{sphere} \mathrm{A}=\mathrm{charge} \mathrm{on} \mathrm{sphere} \mathrm{B}=6.5\times {10}^{-7} \mathrm{C} \\ \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{spheres}-\mathrm{r} =50 \mathrm{cm} = 0.5 \mathrm{m} \\ \mathrm{Force} \mathrm{of} \mathrm{repulsion} \mathrm{as} \mathrm{per} \mathrm{Coulomb}\text{'}\mathrm{s} \mathrm{law} \mathrm{is} \\ \mathrm{F}=\mathrm{k}.\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2} \\ =\frac{9\times {10}^9\times (6.5{)}^2\times ({10}^{-7}{)}^2}{(0.5{)}^2} \\ =1.52\times {10}^{-2} \mathrm{N} \\ \mathrm{a}) \mathrm{if} \mathrm{each} \mathrm{sphere} \mathrm{is} \mathrm{charged} \mathrm{double} \mathrm{and} \mathrm{the} \mathrm{distance} \mathrm{between} \mathrm{them} \mathrm{is} \mathrm{halved} \\ \mathrm{Charge} \mathrm{on} \mathrm{sphere} \mathrm{A}=\mathrm{Charge} \mathrm{on} \mathrm{sphere} \mathrm{B}=2\times (6.5\times {10}^{-7})=13\times {10}^{-7} \\ \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{spheres} - \mathrm{r}=0.25 \mathrm{m} \\ \mathrm{Force} \mathrm{of} \mathrm{repulsion} -\mathrm{F}=\mathrm{k}.\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2} \\ =\frac{9\times {10}^9\times 4\times (6.5{)}^2\times ({10}^{-7}{)}^2}{(0.25{)}^2} \\ = 0.243 \mathrm{N} \end{gathered}$$

b) the two spheres are placed in water of dielectric constant 80 i.e, k=80
$$\begin{gathered} \mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }\mathrm{K}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}=\frac{{\mathrm{F}}_{\mathrm{air}}}{\mathrm{K}} \\ =\frac{1.52\times {10}^{-2}}{80} \\ =1.9\times {10}^{-4} \mathrm{N} \end{gathered}$$

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Consider an infinite thin plane sheet of charge having a uniform surface charge density $\mathrm{\sigma }$ on both sides of sheet. By symmetry, it follows that the electric field is perpendicular to the sheet of charge and is directed in outward direction. 
To find: Electric field due to plane sheet of charge at any point P distant r from it.



We will draw a cylinder of cross-section area A through the point P as gaussian surface (fig. above).
Electric lines of force are parallel to the curved surface of the cylinder therefore the flux due to electric field of plane sheet of charge passes only through the two circular cross section of cylinder.
If E is the magnitude of electric field at point P, then electric flux crossing through the gaussian surface is

     $$\begin{gathered} \mathrm{\varphi }=\stackrel{\rightharpoonup }{\mathrm{E}}\times \mathrm{area} \mathrm{of} \mathrm{end} \mathrm{faces} (\mathrm{circular} \mathrm{cross} \mathrm{section}) \mathrm{of} \mathrm{the} \mathrm{cylinder} \\ \mathrm{\varphi }=\mathrm{E}\times 2\mathrm{A} \\ \mathrm{According} \mathrm{to} \mathrm{Gauss} \mathrm{theorem}, \mathrm{we} \mathrm{have} \\ \mathrm{\varphi }=\frac{\mathrm{q} }{{\mathrm{\epsilon }}_{\circ }} \\ \mathrm{Here}, \\ \mathrm{the} \mathrm{charge} \mathrm{enclosed} \mathrm{by} \mathrm{the} \mathrm{gaussian} \mathrm{surface} \mathrm{is}-\mathrm{q} = \mathrm{\sigma A} \\ \Rightarrow \mathrm{\varphi }=\frac{\mathrm{\sigma A}}{{\mathrm{\epsilon }}_{\circ }} \\ \Rightarrow \mathrm{E}\times 2\mathrm{A}=\frac{\mathrm{\sigma A}}{{\mathrm{\epsilon }}_{\circ }} \\ \Rightarrow \mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_{\circ }} \end{gathered}$$ 
Thus, the magnitude of electric field at a point due to an infinite plane sheet of charge is independent of its distance from the sheet of charge.

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Let, the midpoint of the line joining the two charges be O. 

a) Electric field at O due to charge at A-E_{A=$$\begin{gathered} \frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}.\frac{q_A}{r^2}=\frac{9\times {10}^9\times 2\times {10}^{-6}}{0.1^2} N C^{-1} \\ =1800\times {10}^3 N{C}^{-1} \end{gathered}$$}

Electric field at O due to charge at B-E_{B=$$\begin{gathered} \frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}.\frac{{\mathrm{q}}_{\mathrm{B}}}{{\mathrm{r}}^2}=\frac{9\times {10}^9\times -3\times {10}^{-12}}{0.1^2} \mathrm{N} {\mathrm{C}}^{-1} \\ =2700\times {10}^{-3} {\mathrm{NC}}^{-1} \end{gathered}$$}


Resultant intensity -E =E_A+ E_B                             
                                       
                                         =$4.5 \times {10}^{3 }{\mathrm{NC}}^{-1}$ 

b) If a test charge of magnitude $1.5\times {10}^{-9} \mathrm{C}$ is placed at the centre, then force experienced is ${\mathrm{q}}_{\circ }\mathrm{E}$
                                           
                                               $$\begin{gathered} 1.5\times {10}^{-9}\times 4.5\times {10}^3 \\ =6.75\times {10}^{-6} \mathrm{C} \end{gathered}$$

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Consider a thin spherical shell of radius 'R' with centre O. Let, a charge +q be distributed uniformly on the surface of the shell. 
According to gaussian theorem, 

          $\oint E.ds = \oint \stackrel{\rightharpoonup }{E}. \stackrel{⏜}{n}.ds = \frac{q}{{\epsilon }_o}$ 
Surface area of the sphere = 4$\mathrm{\pi }$r² 

$\therefore \mathrm{E}. 4{\mathrm{\pi r}}^2 = \frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$

Now, at a point on the surface area of the shell 
Let, $\mathrm{\sigma }$ be the surface charge density on the shell, 
then, 
             q = 4$\mathrm{\pi }$R² . $\mathrm{\sigma }$ 

$\Rightarrow$  $\mathrm{E} = \frac{4{\mathrm{\pi R}}^2 \mathrm{\sigma }}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2} = \frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_{\circ }}$ 

Therefore, 
                   $\mathrm{E} = \frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_{\mathrm{o}}}$ 
which is the required electric field at the surface of the shell.

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Given, q₁= +0.2 C
         q₂= +0.4 C
distance between the charges-d=0.1 m 

a)Electric field at the mid point between these two charges:



Electric field due to q₁-E₁=$\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }} \frac{0.2}{0.{05}^2}$
          $=9\times {10}^9\times \frac{0.2}{0.{05}^2}=720\times {10}^{9 } \mathrm{N}$/C
Electric field due to q2-E₂=$\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }} \frac{0.4}{0.{05}^2}$
          $=9\times {10}^9\times \frac{0.4}{0.{05}^2}=1440\times {10}^{9 } \mathrm{N}$/C

Resultant Electric field at mid-point-E=${\stackrel{\rightharpoonup }{\mathrm{E}}}_1+{\stackrel{\rightharpoonup }{E}}_2$
Since the net electric field is acting in opposite direction we have E= 1440$\times {10}^9-720\times {10}^{9 } \mathrm{N}$
                           = $720\times {10}^{9 } \mathrm{N}$/C

b) 

Let point P be on the line joining the charges such that it is 0.05m away from q₂ and 0.15 m away from q_1.

Electric field due to q_{1= $9\times {10}^9\times \frac{0.2}{0.{15}^2}=80\times {10}^{9 }N/C$}
Electric field due to q₂=$9\times {10}^9\times \frac{0.4}{0.{05}^2}=1440\times {10}^{9 }N$/C
Since, Electric field is acting in the same direction

Resultant electric field intensity-E=$$\begin{gathered} 80\times {10}^9 + 1440\times {10}^9 \\ =1520\times {10}^9 \\ =15.2\times {10}^{11 }\mathrm{N}/\mathrm{C} \end{gathered}$$