Given below are two statements:

Statement I : The electric force changes the speed of the charged particle and hence charges its kinetic energy; whereas the magnetic force does not change like kinetic energy of the charged particle.

Statement II : The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerated the moving charged particle along the direction of magnetic field.

In the light of the above statements, chose the most appropriate answer from the option given below

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

$\frac{qE}{m}=4$      

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

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

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

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

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

Class 12 physics chapter 1 Electric Charges and Fields is an important chapter for the CBSE Board exam. In unit 1 of the CBSE Board exam of class 12 Physics, there are questions from three chapters including - Electric Charges and Fields, Electrostatics, and Electrostatic Potential and Capacitance.

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   $\frac{dF}{dx}=0,so$

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