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Question Number 212593 by mr W last updated on 18/Oct/24

Commented by mr W last updated on 18/Oct/24

$a m o r e g e n e r a l c a s e f r o m Q 212514$
	Answered by mr W last updated on 19/Oct/24

	Commented by mr W last updated on 20/Oct/24
	$F=((kq_1 q_2 )/c^2 ) with k=Coulomb constant c^2 =a^2 +b^2 −2ab cos θ c_1 =((m_2 c)/(m_1 +m_2 )), c_2 =((m_1 c)/(m_1 +m_2 )) c_1 b^2 +c_2 a^2 =c(h^2 +c_1 c_2 ) h^2 =((m_1 a^2 +m_2 b^2 )/(m_1 +m_2 ))−((m_1 m_2 c^2 )/((m_1 +m_2 )^2 )) (F/(m_1 g))=(c_1 /h), or using (F/(m_2 g))=(c_2 /h) ((kq_1 q_2 )/(m_1 gc^2 ))=((m_2 c)/((m_1 +m_2 )(√(((m_1 a^2 +m_2 b^2 )/(m_1 +m_2 ))−((m_1 m_2 c^2 )/((m_1 +m_2 )^2 )))))) ((kq_1 q_2 (m_1 +m_2 ))/(m_1 m_2 g))=μ=(c^3 /( (√(((m_1 a^2 +m_2 b^2 )/(m_1 +m_2 ))−((m_1 m_2 c^2 )/((m_1 +m_2 )^2 )))))) μ^2 =(c^6 /( ((m_1 a^2 +m_2 b^2 )/(m_1 +m_2 ))−((m_1 m_2 c^2 )/((m_1 +m_2 )^2 )))) c^6 +((μ^2 m_1 m_2 )/((m_1 +m_2 )^2 ))c^2 −((μ^2 (m_1 a^2 +m_2 b^2 ))/(m_1 +m_2 ))=0 ⇒c^2 =(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))+m_1 a^2 +m_2 b^2 ]))^(1/3) −(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))−m_1 a^2 −m_2 b^2 ]))^(1/3) ⇒a^2 +b^2 −2ab cos θ=(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))+m_1 a^2 +m_2 b^2 ]))^(1/3) −(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))−m_1 a^2 −m_2 b^2 ]))^(1/3) ⇒θ=cos^(−1) {((a^2 +b^2 −(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))+m_1 a^2 +m_2 b^2 ]))^(1/3) +(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))−m_1 a^2 −m_2 b^2 ]))^(1/3) )/(2ab))}$
	$F = \frac{{k q}_{1} q_{2}}{c^{2}} w i t h k = C o u l o m b c o n s t a n t$ $c^{2} = a^{2} + b^{2} - 2 a b \cos θ$ $c_{1} = \frac{m_{2} c}{m_{1} + m_{2}}, c_{2} = \frac{m_{1} c}{m_{1} + m_{2}}$ $c_{1} b^{2} + c_{2} a^{2} = c (h^{2} + c_{1} c_{2})$ $h^{2} = \frac{m_{1} a^{2} + m_{2} b^{2}}{m_{1} + m_{2}} - \frac{m_{1} m_{2} c^{2}}{{(m_{1} + m_{2})}^{2}}$ $\frac{F}{m_{1} g} = \frac{c_{1}}{h}, o r u s i n g \frac{F}{m_{2} g} = \frac{c_{2}}{h}$ $\frac{{k q}_{1} q_{2}}{m_{1} {g c}^{2}} = \frac{m_{2} c}{(m_{1} + m_{2}) \sqrt{\frac{m_{1} a^{2} + m_{2} b^{2}}{m_{1} + m_{2}} - \frac{m_{1} m_{2} c^{2}}{{(m_{1} + m_{2})}^{2}}}}$ $\frac{{k q}_{1} q_{2} (m_{1} + m_{2})}{m_{1} m_{2} g} = μ = \frac{c^{3}}{\sqrt{\frac{m_{1} a^{2} + m_{2} b^{2}}{m_{1} + m_{2}} - \frac{m_{1} m_{2} c^{2}}{{(m_{1} + m_{2})}^{2}}}}$ $μ^{2} = \frac{c^{6}}{\frac{m_{1} a^{2} + m_{2} b^{2}}{m_{1} + m_{2}} - \frac{m_{1} m_{2} c^{2}}{{(m_{1} + m_{2})}^{2}}}$ $c^{6} + \frac{μ^{2} m_{1} m_{2}}{{(m_{1} + m_{2})}^{2}} c^{2} - \frac{μ^{2} (m_{1} a^{2} + m_{2} b^{2})}{m_{1} + m_{2}} = 0$ $\Rightarrow c^{2} = \sqrt[3]{\frac{μ^{2}}{2 (m_{1} + m_{2})} [\sqrt{{(m_{1} a^{2} + m_{2} b^{2})}^{2} + \frac{4 μ^{2} m_{1}^{3} m_{2}^{3}}{27 {(m_{1} + m_{2})}^{4}}} + m_{1} a^{2} + m_{2} b^{2}]} - \sqrt[3]{\frac{μ^{2}}{2 (m_{1} + m_{2})} [\sqrt{{(m_{1} a^{2} + m_{2} b^{2})}^{2} + \frac{4 μ^{2} m_{1}^{3} m_{2}^{3}}{27 {(m_{1} + m_{2})}^{4}}} - m_{1} a^{2} - m_{2} b^{2}]}$ $\Rightarrow a^{2} + b^{2} - 2 a b \cos θ = \sqrt[3]{\frac{μ^{2}}{2 (m_{1} + m_{2})} [\sqrt{{(m_{1} a^{2} + m_{2} b^{2})}^{2} + \frac{4 μ^{2} m_{1}^{3} m_{2}^{3}}{27 {(m_{1} + m_{2})}^{4}}} + m_{1} a^{2} + m_{2} b^{2}]} - \sqrt[3]{\frac{μ^{2}}{2 (m_{1} + m_{2})} [\sqrt{{(m_{1} a^{2} + m_{2} b^{2})}^{2} + \frac{4 μ^{2} m_{1}^{3} m_{2}^{3}}{27 {(m_{1} + m_{2})}^{4}}} - m_{1} a^{2} - m_{2} b^{2}]}$ $\Rightarrow θ = \cos^{- 1} {\frac{a^{2} + b^{2} - \sqrt[3]{\frac{μ^{2}}{2 (m_{1} + m_{2})} [\sqrt{{(m_{1} a^{2} + m_{2} b^{2})}^{2} + \frac{4 μ^{2} m_{1}^{3} m_{2}^{3}}{27 {(m_{1} + m_{2})}^{4}}} + m_{1} a^{2} + m_{2} b^{2}]} + \sqrt[3]{\frac{μ^{2}}{2 (m_{1} + m_{2})} [\sqrt{{(m_{1} a^{2} + m_{2} b^{2})}^{2} + \frac{4 μ^{2} m_{1}^{3} m_{2}^{3}}{27 {(m_{1} + m_{2})}^{4}}} - m_{1} a^{2} - m_{2} b^{2}]}}{2 a b}}$
	Commented by ajfour last updated on 20/Oct/24

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