2-plane-parallel-conducting-plates-are-held-horizontal-one-above-the-other-in-a-vacuum-Electrons-having-a-speed-of-6-10-6-m-s-and-moves-normally-to-the-plates-enter-the-region-between-them-through-a-h

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Question Number 55431 by necx1 last updated on 24/Feb/19

2 plane parallel conducting plates are held horizontal,one above the other in a vacuum.Electrons having a speed of 6×10^6 m/s and moves normally to the plates enter the region between them through a hole in the lower plate which is earthed.What potential must be applied to the other plate so that the electrons just fails to reach it?What is the subsequent motion of these electrons? (ratio of charge to mass of electron= 1.8×10^(11) Ckg^(−1) )

2 p l a n e p a r a l l e l c o n d u c t i n g p l a t e s a r e

h e l d h o r i z o n t a l, o n e a b o v e t h e o t h e r i n

a v a c u u m . E l e c t r o n s h a v i n g a s p e e d o f

6 \times 10^{6} m / s a n d m o v e s n o r m a l l y t o t h e

p l a t e s e n t e r t h e r e g i o n b e t w e e n t h e m

t h r o u g h a h o l e i n t h e l o w e r p l a t e w h i c h

i s e a r t h e d . W h a t p o t e n t i a l m u s t b e

a p p l i e d t o t h e o t h e r p l a t e s o t h a t t h e

e l e c t r o n s j u s t f a i l s t o r e a c h i t ? W h a t

i s t h e s u b s e q u e n t m o t i o n o f t h e s e

e l e c t r o n s ?

(r a t i o o f c h a r g e t o m a s s o f e l e c t r o n =

1.8 \times 10^{11} {C k g}^{- 1})

Commented by necx1 last updated on 24/Feb/19

u r g e n t l y n e e d e d p l e a s e .

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Feb/19

potential lower plate =0 potential upper plate=v eEd=(1/2)mv_(velocity) ^2 E=((v−0)/d) ev=(1/2)mv_(vel) ^2 v=(1/2)×(1/(((e/m))))×v_(vel) ^2 v=(1/2)×(1/(1.8×10^(11) ))×36×10^(12) v=((10)/(36))×((36×10)/1)=100 volt pls check

p o t e n t i a l l o w e r p l a t e = 0

p o t e n t i a l u p p e r p l a t e = v

e E d = \frac{1}{2} {m v}_{v e l o c i t y}^{2}

E = \frac{v - 0}{d}

e v = \frac{1}{2} {m v}_{v e l}^{2}

v = \frac{1}{2} \times \frac{1}{(\frac{e}{m})} \times v_{v e l}^{2}

v = \frac{1}{2} \times \frac{1}{1.8 \times 10^{11}} \times 36 \times 10^{12}

v = \frac{10}{36} \times \frac{36 \times 10}{1} = 100 v o l t

p l s c h e c k

Commented by necx1 last updated on 24/Feb/19

v e r y c o r r e c t . T h a n k y o u s i r .

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