derive electric potential at a distance r from a point charge q

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$L e t u s d e r i v e t h e e l e c t r i c p o t e n t i a l d u e t o a p o s i t i v e c h a r g e Q a t a d i s \tan c e r f r o m i t : L e t u s s h i f t a u n i t p o s i t i v e c h a r g e f r o m t h e p o i n t w h e r e p o t e n t i a l i s r e q u i r e d t o i n f i n i t y a n d f i n d t h e w o r k d o n e b y t h e f o r c e a p p l i e d b y t h e c h a r g e Q o n t h e u n i t c h a r g e . L e t a t a n y i n s \tan t o f t i m e d i s \tan c e o f t h e u n i t c h a r g e f r o m t h e c h a r g e Q i s x t h a n t h e f o r c e a c t i n g o n i t a t t h i s i n s \tan t i s F = \frac{1}{4 π \in} \frac{Q}{x^{2}} L e t u s d i s p l a c e t h e u n i t c h a r g e b y s m a l l d i s \tan c e d x t h i s f o r c e i s r e p u l s i v e a n d a c t i n t h e d i r e c t i o n o f d i s p l a c e m e n t s o w o r k d o n e d W = F d x d W = \frac{1}{4 π \in} \frac{Q}{x^{2}} d x \int d W = \int_{r}^{\infty} \frac{1}{4 π \in} \frac{Q}{x^{2}} d x W = \frac{Q}{4 π \in} \int_{r}^{\infty} \frac{1}{x^{2}} d x = \frac{Q}{4 π \in} (\frac{- 1}{x} {)^{\infty}}_{r} = \frac{- 1 Q}{4 π \in} (\frac{1}{\infty} - \frac{1}{r}) = \frac{1}{4 π \in} \frac{Q}{r} f r o m t h e d e f i n a t i o n o f p o t e n t i a l t h e w o r k d o n e i s e q u a l t o t h e p o t e n t i a l a t t h e p o i n t V = \frac{1}{4 π \in} \frac{Q}{r} R e g a r d s$

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