Answer to Question #177784 in Field Theory for Charles

Question

four equal point charge of +3.0Cμ are placed in air at the four corners of a square that is 40cm apart on a side. Find the force on any of the charge

Answer 1

Let's denote the corners of the square as ABCD (starting from the upper right corner). Let's the side of the square equals 0.4 meters. Let's find the electric force on charge $q_A$ (which is located at corner A) due to other charges. There are three electric forces that acts on charge $q_A$: the force $F_{BA}$ directed upward, the force $F_{DA}$ directed rightward and the force $F_{CA}$ directed along the square diagonal at an angle of $45^{\circ}$. Let's find the $x$- and $y$-components of this forces:

$$F_{BA,x}=0, F_{BA,y}=\dfrac{kq_Bq_A}{r_{BA}^2},$$ $$F_{BA,y}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(3\cdot10^{-6}\ C)^2}{(0.4\ m)^2}=0.51\ N,$$ $$F_{DA,x}=\dfrac{kq_Dq_A}{r_{DA}^2}, F_{DA,y}=0,$$ $$F_{DA,x}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(3\cdot10^{-6}\ C)^2}{(0.4\ m)^2}=0.51\ N,$$ $$F_{CA,x}=\dfrac{kq_Cq_A}{r_{CA}^2}cos\theta,$$ $$F_{CA,x}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(3\cdot10^{-6}\ C)^2}{(\sqrt{2}\cdot0.4\ m)^2}\cdot cos45^{\circ}=0.18\ N,$$ $$F_{CA,y}=\dfrac{kq_Cq_A}{r_{CA}^2}sin\theta,$$ $$F_{CA,y}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(3\cdot10^{-6}\ C)^2}{(\sqrt{2}\cdot0.4\ m)^2}\cdot sin45^{\circ}=0.18\ N.$$

Then, we can write $x$- and $y$-components of resulting electric force that act on charge $q_A$:

$$F_x=F_{DA,x}+F_{CA,x}=0.51\ N+0.18\ N=0.69\ N,$$ $$F_y=F_{BA,y}+F_{CA,y}=0.51\ N+0.18\ N=0.69\ N.$$

Finally, we can find the resulting electric force that act on charge $q_A$ from the Pythagorean theorem:

$$F=\sqrt{F_x^2+F_y^2}=\sqrt{(0.69\ N)^2+(0.69\ N)^2}=0.97\ N.$$