with reference to book i am posting some basic question for students. 1)two charged coducting sphere of radius r_(1 ) and r_2 are positvely charged separated at a distance d from each other. force of interaction is F which is correct answer a)F=(1/(4πε))((q_1 q_2 )/d^2 ) b)F>(1/(4πε))((q_1 q_2 )/d^2 ) c)F<(1/(4πε))((q_1 q_2 )/d^2 ) d)none of these 2)same question as 1 but one is +ve q_1 and another is −ve q_2 then which option is correct a)F=F_(columb) b)F>F_(columb) c)F<F_(columb) d)none of these pls give answer with explanatiin..


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Question Number 58614 by tanmay last updated on 26/Apr/19

$w i t h r e f e r e n c e t o b o o k i a m p o s t i n g s o m e b a s i c$ $q u e s t i o n f o r s t u d e n t s .$ $1) t w o c h a r g e d c o d u c t i n g s p h e r e o f r a d i u s$ $r_{1} a n d r_{2} a r e p o s i t v e l y c h a r g e d s e p a r a t e d$ $a t a d i s t a n c e d f r o m e a c h o t h e r .$ $f o r c e o f i n t e r a c t i o n i s F$ $w h i c h i s c o r r e c t a n s w e r$ $a) F = \frac{1}{4 π ϵ} \frac{q_{1} q_{2}}{d^{2}}$ $b) F > \frac{1}{4 π ϵ} \frac{q_{1} q_{2}}{d^{2}}$ $c) F < \frac{1}{4 π ϵ} \frac{q_{1} q_{2}}{d^{2}}$ $d) n o n e o f t h e s e$ $2) s a m e q u e s t i o n a s 1 b u t o n e i s + v e q_{1} a n d$ $a n o t h e r i s - v e q_{2} t h e n w h i c h o p t i o n i s c o r r e c t$ $a) F = F_{c o l u m b}$ $b) F > F_{c o l u m b}$ $c) F < F_{c o l u m b}$ $d) n o n e o f t h e s e$ $p l s g i v e a n s w e r w i t h e x p l a n a t i i n . .$
	Answered by Kunal12588 last updated on 27/Apr/19

	$i t h i n k$ $a n s 1 - (b)$ $a n s 2 - (b)$
	Commented by tanmay last updated on 27/Apr/19

	Answered by sridhar nayak last updated on 09/Jun/19

	$f i r s t q u e s t i o n a n s w e r i s 1 . B e s a u s e f o r c e b e t w e e n t w o c h a r g e d s p h e r e i s e q u a l t o t h a t f o r m u l a .$
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