Question

two charges attract each other with a force of 4 into 10 ^ - 6 Newton when they are 0.4 M apart when their distance of substation increases 0.6 m.find the force between them ​

Answer 1

$\boxed{ \bold{ \sf{ \orange{ \huge{Answer}}}}} \\ \\ \star \sf \: \underline{GIVEN :} \\ \\ \implies \sf \: force \: of \: attraction \: between \: two \: charge \\ \sf \: is \: 4 \times {10}^{ - 6} \: N \: when \: they \: are \: 0.4 \: m \: apart \\ \\ \star \sf \: \underline{TO \: FIND :} \\ \\ \implies \sf \: force \: of \: attraction \: between \: them \: they \\ \sf \: are \: 0.6 \: m \: apart \\ \\ \star \sf \: \underline{FORMULA :} \\ \\ \implies \sf \: electric \: force \: between \: two \: charge \: at \: rest \\ \sf \: is \: given \: as \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{\boxed{ \red{ \bold{ \sf{F = \frac{kq1q2}{ {r}^{2} } }}}}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: \sf \: \underline{ \boxed{ \bold{ \blue{F \propto \: \frac{1}{ {r}^{2} }}}}} \\ \\ \star \sf \: \underline{CALCULATION :} \\ \\ \implies \sf \: \frac{F1}{F2} = \frac{ {r2}^{2} }{ {r1}^{2} } \\ \\ \implies \sf \: \frac{4 \times {10}^{ - 6} }{F2} = \frac{ {(0.6)}^{2} }{ {(0.4)}^{2} } \\ \\ \implies \sf \: \underline{\boxed{ \bold{ \green{F2 = 1.78 \times {10}^{ - 6} \: N}}}}$

Answer 2

Explanation:

F2= 1.78×10^6 the force between them