Answer to Question #98781 in Electricity and Magnetism for Caleb

Question

Two equal positive charges q1=q2=2.0C are located at x=0, y=0.30m and x=0, y= - 0.30m, respectively. What are the magnitude and direction of the total electric force that q1 and q2 exert on the third charge Q=4.0C at x=0.40m, y=0?

Answer 1

Define r₁

$r_1=\sqrt{0.3^2+0.4^2}=\sqrt{0.09+0.16}=\sqrt{0.25}=0.5(m)$

Define r₂

$r_2=\sqrt{0.3^2+0.4^2}=\sqrt{0.09+0.16}=\sqrt{0.25}=0.5(m)$

Define the angle $\alpha$

$\alpha=\arccos({0.4/0.5})=36,86^0$

Define Strength $F_1$

$F_1=\frac{1}{4\pi\epsilon_0}\cdot\frac{g_1Q}{r_1^2}=\frac{1}{4\pi\cdot8,85\cdot 10^{-12}(F/m)}\cdot\frac{2(C)\cdot 4(C)}{0.5^2(m^2)}=2,877\cdot10^{11}(N)$

from tog $q_1=q_2$ and $r_1=r_2$ follows that $F_2=F_1$

then

$F=2\cdot F_1 \cdot \cos(\alpha)=2\cdot 2,877 \cdot 10^{11}\cdot \cos(36,86^0)=4,603 \cdot 10^{11}(N)$