1.34) Suppose that the particle in is an electron projected with velocity v_{x}=2.0 \times 10^{6}ms^{-1}. If E between the plates separated by 0.5 cm is 9.1 \times 10^{2}\frac{N}{C}, where will the electron strike the upper plate? (|e|=\left | e \right |=1.6 \times 10^{-19}, m_{e}=9.1\times 10^{-31}kg)

Answers (1)
S safeer

\therefore The vertical deflection of the particle at the far edge of the plate is s=\frac{qEL^{2}}{2mv_{x}^{2}}

given s= 0.5cm=0.005cm

calculate for L from the above equation 

L=\sqrt{\frac{2smv_x^2}{qE}}=\sqrt{\frac{2\times 0.005\times 9.1\times 10^{-31}(2\times10^6)^2 }{1.6\times 10^{-19}\times9.1\times10^2}}

L=1.6 cm

 

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Foundation 2021 Class 9th Maths

Master Maths with "Foundation course for class 9th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions