Problem — Resultant force on three charges

Problem

Three point charges in vacuum: $q_1 = +2\ \mu\text{C}$ at $A=(0;\,0)$, $q_2 = +3\ \mu\text{C}$ at $B=(0{,}10;\,0)\ \text{m}$, $q_3 = -1\ \mu\text{C}$ at $C=(0{,}05;\,0)\ \text{m}$. Calculate the resultant force on $q_3$, giving magnitude and direction along the $x$ axis.

Solution

Use $k = 8{,}99\cdot 10^9\ \text{N·m}^2/\text{C}^2$. Charge $q_3$ sits at the midpoint, at distance $r = 0{,}05\ \text{m}$ from both $q_1$ and $q_2$. Apply the superposition principle: sum the two Coulomb forces along the $x$ axis.

Force of $q_1$ on $q_3$ ($q_1>0$, $q_3<0$: attractive, pulls $q_3$ towards $-x$):

$F_{13} = k\,\frac{|q_1|\,|q_3|}{r^2} = 8{,}99\cdot 10^9 \cdot \frac{(2\cdot 10^{-6})(1\cdot 10^{-6})}{(0{,}05)^2} = \frac{8{,}99\cdot 10^9 \cdot 2\cdot 10^{-12}}{2{,}5\cdot 10^{-3}} \approx 7{,}19\ \text{N}$

Force of $q_2$ on $q_3$ ($q_2>0$, $q_3<0$: attractive, pulls $q_3$ towards $+x$):

$F_{23} = k\,\frac{|q_2|\,|q_3|}{r^2} = 8{,}99\cdot 10^9 \cdot \frac{(3\cdot 10^{-6})(1\cdot 10^{-6})}{(0{,}05)^2} = \frac{8{,}99\cdot 10^9 \cdot 3\cdot 10^{-12}}{2{,}5\cdot 10^{-3}} \approx 10{,}79\ \text{N}$

Vector sum (the two forces are opposite along $x$):

$F_x = F_{23} - F_{13} = 10{,}79 - 7{,}19 = 3{,}6\ \text{N} \quad (\text{towards } +x)$

$\ev{\vv{F}\approx(3{,}6;\,0)\ \text{N}\ \text{towards }+x}$

Links

Topics: Elettrostatica Concepts: Legge di Coulomb Skills: Principio di sovrapposizione