The force between two charges 0.06,m apart is | vedclass

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(b) $F \propto \frac{1}{{{r^2}}} \Rightarrow \frac{{{F_1}}}{{{F_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2}$==>$\frac{5}{{{F_2}}} = {\left(

Vedclass · Accepted Answer

$11.25$

Vedclass · Answer

(b) $F \propto \frac{1}{{{r^2}}} \Rightarrow \frac{{{F_1}}}{{{F_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2}$==>$\frac{5}{{{F_2}}} = {\left( {\frac{{0.04}}{{0.06}}} \right)^2} = {F_2} = 11.25\,N$

The force between two charges $0.06\,m$ apart is $5\,N$. If each charge is moved towards the other by $0.01\,m$, then the force between them will become.........$N$

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