Question

Two small beads having positive charges 3q and q are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to the point $x=d$. As shown in figure, a third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium?

• A. $\frac{\sqrt{3}d}{(1+\sqrt{3})}$ from +3q charge, yes
• B. $\frac{\sqrt{5}d}{(1+\sqrt{3})}$ from +3q charge, yes
• C. $\frac{\sqrt{2}d}{(1+\sqrt{3})}$ from +3q charge, yes
• D. $\frac{\sqrt{3}d}{(1+\sqrt{2})}$ from +3q charge, yes

Answer 1

Answer: (A)

Let the third charge Q is at position x from charge 3q and $(d-x)$ from q.

Now Q to be in equilibrium , if $\frac{k\left(3q\right)Q}{x^2}=\frac{kqQ}{(d-x{)}^2}$

or $3(d^2-2xd+x^2)=x^2$

or $2x^2-6xd+3d^2=0$

or $x=\frac{6d\pm \sqrt{36d^2-24d^2}}{4}=\frac{d}{2}(3\pm \sqrt{3})$

The equilibrium point should be in between 0 to d so $x=\frac{d}{2}(3-\sqrt{3})=\frac{d}{2}\frac{(3-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})}=\frac{d}{2}\frac{9-3}{(3+\sqrt{3})}=\frac{\sqrt{3}d}{(\sqrt{3}+1)}$

The bead can be in stable equilibrium, if it has positive charge.