Answer to Question #294694 in Optics for sarah

Question

In the figure above, q₁ = 6.53 nC, q₂ = 13.1 nC, and q₃ = 13.1 nC. If the net force on q₁ is 0, then what would q be?

Answer 1

Force

$F_3=\frac{kq_1q_3}{a^2}=\frac{9\times10^9\times13.1\times10^{-9}\times q}{a^2}$

$F_2=\frac{kq_1q_2}{a^2}=\frac{9\times10^9\times13.1\times10^{-9}\times q}{a^2}$

Net force

$F_{net}=\sqrt{2}F_2$

$F_1=\frac{kq_1q_2}{a^2}=\frac{9\times10^9\times6.53\times10^{-9}\times q }{2a^2}$

$F'_{net}=\sqrt{2}\frac{kq_1q_2}{a^2}=\sqrt{2}\times\frac{9\times10^9\times6.53\times10^{-9}\times13.1\times10^{-9}}{a^2}$

$F_{net}=({\sqrt{2}\times6.53+\frac{1}{2}\times q\times10^{-9}})\times13.1\times10^{-18}$

$F_{net}=0$

$({\sqrt{2}\times6.53\times10^{-9}+\frac{1}{2}\times q})=0$

$q=-18.46\times10^{-9}C$