Two identical metal spheres gallery charges + q and minus 2 respectively when the spheres are separated by a large distance are the force of the f between them is now the spheres are allowed to touch and then move back to the same separation find the new force of repulsion between them

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Solution

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$T h e f o r c e b e t w e e n + q a n d - 2 q a t a d i s t a n c e r i s g i v e n a s F N o w F = \frac{K ((+ q) \times (- 2 q))}{r^{2}} = \frac{- 2 k q^{2}}{r^{2}} W h e n t h e c h a g e s + q a n d - 2 q c o m e s i n c o n t a c t, t h e n i t n e u t r a l i s e s t h e c h a g e + q b y - q c h a r g e a n d l e f t o n l y - q . T h e s e - q c h a r g e w i l l b e e q u a l l y d i s t r i b u t e d b o t h t h e s p h a r e . I f i t i s s e p e r a t e d t h e t w o s p h e r e a g a i n t h e n e a c h s p h e r e w i l l a q u i r e a m o u n t o f c h a r g e o f \frac{- q}{2} a n d \frac{- q}{2} . N o w t h e n e w f o r c e b e t w e e n t h e m w i l l b e F_{n} = K \frac{((\frac{- q}{2})^{2})}{2} = \frac{1}{4} ((\frac{k q^{2}}{2})) = \frac{(- F)}{8} S o t h e f o r c e w i l l b e 1 / 8 p a r t o f p r e v i o u s f o r c e a n d i t w i l l b e r e p u l s i v e f o c e w h i l e F w a s a t t r a c t i v e .$

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