What is the weightage of Electric Charges and Fields in CBSE Board exam?

From Gauss law

$?=\frac{q_{enclosed}}{{\epsilon }_0}=\frac{5q}{{\epsilon }_0}$

For maximum value of s, initially, electron must move away from plate.

ut + $\frac{1}{2}$ at² = s

t = 1u = 1m/s               s = –1 m

1 × 1 – $\frac{1}{2}$ a × 1² = – 1

-> a = 4m/s²

$\frac{qE}{m}=4$      

  $\frac{q\sigma }{2{\epsilon }_{0}m}=4$     

$\sigma =\frac{4\times 2\times 9\times 10^{-12}\times 9\times 10^{-31}}{1.6\times 10^{-19}}$

$=\frac{8\times 81}{1.6}\times 10^{-24}$      

= 4.05 × 10^–22C/m²

${?}_1=\frac{3}{5}{E}_0\left(0.2\right)N{m}^2{C}^{-1},and{?}_2=\frac{4}{5}{E}_0\left(0.3\right)N{m}^2{C}^{-1}$

$\Rightarrow \frac{{?}_1}{{?}_2}=\frac{3\times 0.2\times 5}{5\times 0.3\times 4}=\frac{1}{2}$

According to question, we can write

    $F_{CA}=F_{CB}=\frac{KqQ}{x^2+{\left(\frac{d}{2}\right)}^2}\Rightarrow F=2F_{CA}cos\theta =\frac{2KqQx}{{\left[x^2+{\left(\frac{d}{2}\right)}^2\right]}^{\frac{3}{2}}}$                    

For maxima of force   $\frac{dF}{dx}=0,so$

$x=\frac{d}{2\sqrt[]{2}}$

In y axis u? = v? a? = -E? q / m
s? = 0, u? t + (1/2)a? t² = 0 ⇒ t = 2u? /a? = 2v? m/E? q
x coordinate at that time = v? × t = (2v? m/0) × v? = (2v? ²m)/E? q