prosze o rozwiązanie tego zadanka - elektrostatyka;p dane; r=6*10^-5 F=4kN q1= 2*10^-5 Culomba Szukane; q2=? F=k*(q1*q2)/r^2

Dane:   [latex]F=4kN=4000N[/latex]   [latex]r=6*106*10^{-5}[/latex]   [latex]k=frac{1}{4}[/latex]   [latex]q_{1}=2*10^{-5}[/latex]     Szukane:   [latex]q_{2}=?[/latex]     Rozwiązanie:   [latex]F=kfrac{q_{1}*q_{2}}{r^{2}}||*r^{2}[/latex]   [latex]F*r^{2}=k*q_{1}*q_{2}||/k*q_{1}[/latex]   [latex]frac{F*r^{2}}{k*q_{1}}=q_{2}||<-->[/latex]   [latex]q_{2}=frac{F*r^{2}}{k*q_{1}}[/latex]   [latex]q_{2}=frac{4000N*6*10^{-10}}{frac{1}{4}*{2*10^{-5}}}[/latex]   [latex]q_{2}=frac{0,0000024}{0,00005}[/latex]   [latex]q_{2}=0,048[C][/latex].