Question

Two positive charges, each of 4.8 µC, and negative charge -6.36 µC are fixed at the vertices of an equilateral triangle of side 13 cm. Find the electrical force on negative charge.

Answer 1

We are given

$q_1=q_2=q=+4.8 \cdot 10^{-6}C$

$q_3=-6.36 \cdot 10^{-6}C$

$r=0.13m$

Force acting on charge q3 from the side of charge q1

$F_{13}=\frac{1}{4 \pi\epsilon_0 } \cdot \frac{q_1 \cdot q_3}{r^2}=\frac{1}{4 \pi\epsilon_0 } \cdot \frac{q \cdot q_3}{r^2}$

Force acting on charge q3 from the side of charge q2

$F_{23}=\frac{1}{4 \pi\epsilon_0 } \cdot \frac{q_2 \cdot q_3}{r^2}=\frac{1}{4 \pi\epsilon_0 } \cdot \frac{q \cdot q_3}{r^2}$

$F_{13}=F_{23}=F_0$

Then behind the cosine theorem we write

$F^2=F_{13}^2+F_{23}^2-2 \cdot F_{13} \cdot F_{23} \cdot \cos(\gamma)$

$F^2=F_{13}^2+F_{23}^2-2 \cdot F_{13} \cdot F_{23} \cdot \cos(180^0-2 \alpha)$

$F^2=2F_{0}^2+2 F_{0}^2 \cdot \cos(2 \alpha)=2F_0^2 \cdot (1+\cos(60^0))$

$F^2=2F_0^2 \cdot (1+\frac{1}{2})=2F_0^2 \cdot \frac{3}{2}=3 \cdot F_0^2$

$F=\sqrt {3} \cdot F_0=\sqrt {3} \cdot\frac{1}{4 \pi\epsilon_0 } \cdot \frac{q \cdot q_3}{r^2}$

$F=\sqrt {3} \cdot\frac{1}{4 \pi\cdot 8.854 \cdot 10^{-12} } \cdot \frac{4.8 \cdot 10^{-6} \cdot 6.36 \cdot 10^{-6}}{0.13^2}=28.12N$