zad. 5 Daję naj Proszę o wytłumaczenie

prawo Coulomba : [latex]F = frac{kq_1q_2}{r^2} [/latex] k = 9 *10⁹ N * m²/C² obl. dla kolejnych wierszy : 1. q₁ = 8 μC = 8 *10⁻⁶ C q₂ = 10 μC = 10 *10⁻⁶ C F = 72 N  r = ? [latex]F = frac{kq_1q_2}{r^2} /*r^2 F*r^2 = kq_1q_2/: F [/latex] [latex]r^2 = frac{kq_1q_2}{F } r = sqrt{ frac{kq_1q_2}{F} } [/latex] [latex]r = sqrt{ frac{9 *10^9 frac{N*m^2}{C^2} * 8 *10 ^{-6} C * 10 ^{-6} C }{72 N} } = 0,1 m [/latex] 2. q_1 = 24 *10⁻⁶ C q₂ = 10 * 10⁻⁶ C F = ? [latex]F = frac{9 *10 ^{9} frac{N*m^2}{C^2}* 24 *10 ^{-6} C * 10 ^{-6 C} }{(10 ^{-2} m)^2 } = 2,1 *10 ^{3} N[/latex] 3. q₁ = 24 *10⁻⁶ C q₂ = 20 *10⁻⁶ C [latex]F = frac{9 *10 ^{9} frac{N*m^2}{C^2} *24 *10 ^{-6} C * 20 *10 ^{-6} C }{(10 ^{-2} m)^2} = 4,32 *10^4 N[/latex] 4. q₁ = 8 *10⁻⁶ C F = 144 N r = 10⁻² m q₂ = ? [latex]F = frac{kq_1q_2}{r^2 } /* r^2 F*r^2 = kq_1q_2 /: kq_1[/latex] [latex]q_2 = frac{F*r^2}{kq_1} [/latex] [latex]q_2 = frac{144 N * ( (10 ^{-2}m)^2 }{9*10^9 frac{N*m^2}{C^2}*8 *10 ^{-6} C } = 2 *10 ^{-7} C[/latex] 5. q₁= ? q₂ = 40 *10⁻⁶ C F = 144 N [latex]q_1 = frac{F*r^2}{kq_2} [/latex] [latex]q_1 = frac{144 N * (10 ^{-2}m)^2 }{9 *10^9 frac{N*m^2}{C^2}* 40 *10 ^{-6} C } = 4 *10 ^{-8} C[/latex] 5. q₁ = 40 *10 ⁻⁶ C q₂ = ? F = 72 N [latex]q_2 = frac{F*r^2}{kq_1} [/latex] [latex]q_2 = frac{72 N * (10 ^{-2} )^2}{9 *10^9 frac{N*m^2}{C^2} * 40 *10 ^{-6} C } = 2 *10 ^{-8} C[/latex]