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. $-(3+\sqrt{8})$ only
• B. $-3+\sqrt{8}$ only
• C. $-(3+\sqrt{8})\,or\ (-3+\sqrt{8})$
• D. $+\sqrt{3}$

Answer 1

Answer: (C)

$-(3+\sqrt{8})\,or\ (-3+\sqrt{8})$

Answer 2

Since the two charges (spheres) attract, they will be opposite in sign,
ie, ${{q}_{1}}$ and $-{{q}_{2}}$ .
Force, $F=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{{{q}_{1}}\times \,-{{q}_{2}}}{{{d}^{2}}}$
After touching, charge on each will be
$\frac{{{q}_{1}}-{{q}_{2}}}{2}$ .
New force, $F=\frac{1}{4\pi {{\varepsilon }_{0}}}{{\left( \frac{{{q}_{1}}-{{q}_{2}}}{2} \right)}^{2}}\times \frac{1}{{{d}^{2}}}$
Given $|F|=|F|$
On solving by quadratic equations, we get
$\frac{{{q}_{1}}}{{{q}_{2}}}=-(3+\sqrt{8})$
or $(-3+\sqrt{8})$