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1. Electric Charges and Fields
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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.........$N$
A
$7.20$
B
$11.25$
C
$22.50$
D
$45$
Solution
(b) $F \propto \frac{1}{{{r^2}}} \Rightarrow \frac{{{F_1}}}{{{F_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2}$==>$\frac{5}{{{F_2}}} = {\left( {\frac{{0.04}}{{0.06}}} \right)^2} = {F_2} = 11.25\,N$
Standard 12
Physics
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