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  • A uniform electric field of 10 N/C is created between two parallel charged pates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5eV. The

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

Option: 1

45   
  
 


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer


               
\begin{aligned} & \therefore F=e E \\ & \Rightarrow a=\frac{e E}{m} \end{aligned}
The electron will take a parabolic path i.e., projectile motion.
Here, S_x  = 10cm = 0.1m 
\therefore t=\frac{0.1}{u_x}----(i)
\begin{aligned} & \text { Now } \mathrm{v}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}+a_y t \\ & \Rightarrow \mathrm{v}_{\mathrm{y}}=0+\frac{e E}{m} \times \frac{0.1}{u_x}---(i i) \end{aligned}
Also \\ K E=\frac{1}{2} m v^2=\frac{1}{2} m u_x^2

m u_x^2=2 \times K E=2 \times 0.5 e=e----(i i i)

\Rightarrow \mathrm{u}_{\mathrm{x}}=\sqrt{\frac{\mathrm{e}}{\mathrm{m}}}

\operatorname{Tan} \theta=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{u}_{\mathrm{x}}}=\frac{\frac{e E}{m} \times \frac{0.1}{u_x}}{u_x}=\frac{e E}{m} \times \frac{0.1}{u^2_x}=E\times 0.1=10\times 0.1=1


\begin{aligned} & \Rightarrow \tan \theta=1 \\ & \therefore \theta=45^{\circ} \end{aligned}
 
 

Posted by

Pankaj Sanodiya

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