Question

Two charges, magnitude + 2 x 10-6 C each are 60cm apart. Find the magnitude of the force exerted by these charges on a third charge of magnitude + 4 x 10-6 C that is 50 cm away from each of the first two charges.

Answer 1

Solution. Since charges of the same sign, two charges repel the third with the forces F₁ and F₂ equal in absolute value

According to the conditions of the problem r=0.6m; r₁=r₂=0.5m; q₁=q₂=2x10^-6С; q₃=4x10^-6C.

$$F_1=F_2=\frac{kq_1q_3}{r_1^2}=\frac{kq_2q_3}{r_2^2}=\frac{9\times10^9\times 2\times 10^{-6}\times 4\times 10^{-6}}{0.5^2}=0.288N$$

The angle between the vectors is equal to the angle ACB. Therefore, the resulting force is

$$\vec{F}=\vec{F_1}+\vec{F_2}$$

Using the cosine theorem the magnitude of the resulting force

$$F^2=F_1^2+F_2^2-2F_1F_2cos(180^0-∠ACB)$$ $$F=\sqrt{F_1^2+F_2^2-2F_1F_2cos(180^0-∠ACB)}=\sqrt{2F_1^2+2F_1^2cos∠ACB}$$

$$F=F_1\sqrt{2+2cos∠ACB}$$

Using the cosine theorem for ∆ABC get

$$r^2=r_1^2+r_2^2-2r_1r_2cos∠ACB$$ $$cos∠ACB=\frac{r_1^2+r_2^2-r^2}{2r_1r_2}=\frac{0.5^2+0.5^2-0.6^2}{2\times0.5\times0.5}=0.28$$

As result

$$F=0.288\sqrt{2+2\times 0.28}=0.4608N$$

Answer. 0.4608N