Two identical conducting spheres carrying different charges attract each other with a force F when placed in air medium at a distance d apart. The spheres are brought into contact and then taken to their original positions. Now the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force. The ratio between initial charges on the spheres is

Question

Two identical conducting spheres carrying different charges attract each other with a force F when placed in air medium at a distance d apart. The spheres are brought into contact and then taken to their original positions. Now the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force. The ratio between initial charges on the spheres is (A) -(.3+√8.) only (B) -3+√8 only (C) -(.3+√8.) or ( -3+√8 ) (D) -√8

Tardigrade Admin · Accepted Answer

F= (k q1 q2/d2) After touching textitF'= (k/( textitd)2)((( textitq)1 - q2/2))2 F =F' q1q2 = ((q1 - q2/2))2 (q1/q2) = - 3 + √8 or - 3 - √8

$- (3 + 8)$ only

$- 3 + 8$ only

$- (3 + 8)$ or ( $- 3 + 8$ )

$- 8$

$F = \frac{k q _{1} q _{2}}{d ^{2}}$
After touching $F’ = \frac{k}{( d ) ^{2}} (\frac{( q ) _{1} - q _{2}}{2})^{2}$
$F = F^{'}$
$q_{1} q_{2} = (\frac{q _{1} - q _{2}}{2})^{2}$
$\frac{q _{1}}{q _{2}} = - 3 + 8 or - 3 - 8$

−(3+8​) only

−3+8​ only

−(3+8​) or ( −3+8​ )

−8​

F=d2kq1​q2​​ After touching F’=(d)2k​(2(q)1​−q2​​)2 F=F′ q1​q2​=(2q1​−q2​​)2 q2​q1​​=−3+8​or−3−8​

$- (3 + 8)$ only

$- 3 + 8$ only

$- (3 + 8)$ or ( $- 3 + 8$ )

$- 8$

$F = \frac{k q _{1} q _{2}}{d ^{2}}$
After touching $F’ = \frac{k}{( d ) ^{2}} (\frac{( q ) _{1} - q _{2}}{2})^{2}$
$F = F^{'}$
$q_{1} q_{2} = (\frac{q _{1} - q _{2}}{2})^{2}$
$\frac{q _{1}}{q _{2}} = - 3 + 8 or - 3 - 8$