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

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

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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