A uniform electric field of 10 N / C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ) of deviation of the path of electron as it comes out of the field is (in degree).

Question

A uniform electric field of 10 N / C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ) of deviation of the path of electron as it comes out of the field is (in degree). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

0.5 e =(1/2) mv x 2 ⇒ v x =√( e / m ) Along x L = v x t =√( e / m ) t Along y v y =( eE / m ) t operatornamedividing ( v y / L )= E √( e / m )= Ev x ⇒ operatornameTan θ=( v y / v x )= E × L =10 × 0.1=1 θ=45°

0.5e=21​mvx2​⇒vx​=me​​ Along xL=vx​t=me​t​ Along yvy​=meE​t dividingLvy​​=Eme​​=Evx​ ⇒Tanθ=vx​vy​​=E×L=10×0.1=1 θ=45∘

$0.5 e = \frac{1}{2} m v_{x}^{2} \Rightarrow v_{x} = \frac{e}{m}$
Along $xL = v_{x} t = \frac{e}{m} t$
Along $y v_{y} = \frac{e E}{m} t$
$dividing \frac{v _{y}}{L} = E \frac{e}{m} = E v_{x}$
$\Rightarrow Tan θ = \frac{v _{y}}{v _{x}} = E \times L = 10 \times 0.1 = 1$
$θ = 4 5^{\circ}$