| Force of repulsion between two charges od \[+\,2\mu C\] and \[+\,3\mu C\]separated by a distance of \[10\,cm{{s}^{-\,1}}\]is \[{{F}_{1}}.\]If a dielectric slab of dielectric constant \[k=4\]is placed, such that it fills 5 cm distance (as shown), then force between charges will be |
 |
A) \[\frac{4}{9}{{F}_{1}}\]
B) \[\frac{9}{4}{{F}_{1}}\]
C) \[\frac{7}{5}{{F}_{1}}\]
D) \[\frac{5}{7}{{F}_{1}}\]
Correct Answer: A
Solution :
| Effective air separation between charges \[=r-t+t\sqrt{k}\] |
| \[=10-5+5\sqrt{4}=20-5=15\,cm\] |
| \[\therefore \] \[{{F}_{1}}\propto \frac{{{q}_{1}}{{q}_{2}}}{{{(10)}^{2}}}\]and \[{{F}_{2}}\propto \frac{{{q}_{1}}{{q}_{2}}}{{{15}^{2}}}\] |
| \[\therefore \] \[\frac{{{F}_{2}}}{{{F}_{1}}}=\frac{{{(10)}^{2}}}{{{(15)}^{2}}}\]\[\Rightarrow \]\[{{F}_{2}}=\frac{4}{9}{{F}_{1}}\] |
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