Chapter 21 - Coulomb's Law - Questions - Page 623: 3

$a$ and $b$

We are trying to find the equilibrium of an electron and the charges at distance left of the particles. This is a solution to the, $F_{q1} = F_{q2}$, equation The solution has to be positive because a negative sign changes the direction of the equilibrium Let r = distance from electron to q1 Let d = distance between the two charges Let e = charge of an electron $a)$ $F_{q1} =k\frac{eq}{r^2}$ $ F_{q2} = k\frac{3eq}{(r+d)^2} $ $F_{q1} = F_{q2}$ $k\frac{eq}{r^2} = k\frac{3eq}{(r+d)^2} $ $(r+d)^2 = 3r^2$ $3r^2 = (r+d)^2$ $\sqrt 3r = r+d$ $\sqrt 3r - r =d$ $r (\sqrt 3 - 1) =d$ $r =\frac{d}{\sqrt 3 - 1}$ The solution for r is positive so the electron will be at equilibrium at the left of the particles $b)$ $F_{q1} =k\frac{eq}{r^2}$ $ F_{q2} = k\frac{3eq}{(r+d)^2} $ $F_{q1} = F_{q2}$ $k\frac{eq}{r^2} = k\frac{3eq}{(r+d)^2} $ $(r+d)^2 = 3r^2$ $3r^2 = (r+d)^2$ $\sqrt 3r = r+d$ $\sqrt 3r - r =d$ $r (\sqrt 3 - 1) =d$ $r =\frac{d}{\sqrt 3 - 1}$ The solution for r is positive so the electron will be at equilibrium at the left of the particles $c)$ $F_{q1} =k\frac{eq}{(r+d)^2}$ $ F_{q2} = k\frac{3eq}{(r^2} $ $F_{q1} = F_{q2}$ $ k\frac{eq}{(r+d)^2} = k\frac{3eq}{(r^2} $ $r^2 = 3(r+d)^2 $ $r = \sqrt 3r+\sqrt 3d) $ $r - \sqrt 3r = \sqrt 3d$ $r(1-\sqrt 3) = \sqrt 3d$ $r = \frac{\sqrt 3d}{1-\sqrt 3}$ $1 -\sqrt 3$ will result in a negative value thereby making the whole result negative $d)$ $F_{q1} =k\frac{eq}{(r+d)^2}$ $ F_{q2} = k\frac{3eq}{(r^2} $ $F_{q1} = F_{q2}$ $ k\frac{eq}{(r+d)^2} = k\frac{3eq}{(r^2} $ $r^2 = 3(r+d)^2 $ $r = \sqrt 3r+\sqrt 3d) $ $r - \sqrt 3r = \sqrt 3d$ $r(1-\sqrt 3) = \sqrt 3d$ $r = \frac{\sqrt 3d}{1-\sqrt 3} $1 -\sqrt 3$ will result in a negative value thereby making the whole result negative Therefore the answer is a and b4