Two large parallel metal plates carrying opposite charges are seperated by a distance of 0.2m,and the potential difference between them is 0.5KV. (a)Calculate the magnitude of the uniform electric field between them. (b)Also calculate the workdone by this field on a charge of 4.0×10^(−9) C when it moves from the plate of higher potential to that of lower potential.


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Question Number 40818 by Necxx last updated on 27/Jul/18

$T w o l a r g e p a r a l l e l m e t a l p l a t e s c a r r y i n g$ $o p p o s i t e c h a r g e s a r e s e p e r a t e d b y a$ $d i s t a n c e o f 0.2 m, a n d t h e p o t e n t i a l$ $d i f f e r e n c e b e t w e e n t h e m i s 0.5 K V .$ $(a) C a l c u l a t e t h e m a g n i t u d e o f t h e$ $u n i f o r m e l e c t r i c f i e l d b e t w e e n t h e m .$ $(b) A l s o c a l c u l a t e t h e w o r k d o n e b y$ $t h i s f i e l d o n a c h a r g e o f 4.0 \times 10^{- 9} C$ $w h e n i t m o v e s f r o m t h e p l a t e o f$ $h i g h e r p o t e n t i a l t o t h a t o f l o w e r$ $p o t e n t i a l .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

	$E = \frac{V}{d} = \frac{0.5 \times 10^{3}}{0.2} = 2.5 \times 10^{3}$ $W = q V = 4.0 \times 10^{- 9} \times 0.5 \times 10^{3} = 2.0 \times 10^{- 6}$
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