Question

Please answer this question with explanation. The force between two charges 0.06m apart is 5N. If each charge is moved towards the other by 0.01m, then the force between them will become a)7.2N b)11.25N c)22.5N d)45N

Answer 1

your answer is 11.25 N
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Answer 2

The force between the charges is a) 7.2 N at 0.01 m

Given:

Force = 5 N

Distance = 0.06 m

Solution:  

Now, if "charge is moved" towards the other by 0.01 m that is the "distance between them" is decreased, so the force will now be increased.  

$F=\frac{k q_{1} q_{2}}{r^{2}}$

We know that,

F = 5 N  

$5=\frac{k q_{1} q_{2}}{(0.06)^{2}}$

$k q_{1} q_{2}=5 \times(0.06)^{2} \rightarrow(1)$

Now when the distance is decreased by 0.01, So the force will be

$F^{\prime}=\frac{k q_{1} q_{2}}{(0.05)^{2}} \rightarrow(2)$

So, from first equation, we have,

The equation (1) will not change, thereby, substituting equation (1) in equation (2), we get,

$F^{\prime}=\frac{5 \times(0.06)^{2}}{(0.05)^{2}}$

$=\frac{0.018}{2.5 \times 10^{-3}}$

$\Rightarrow F^{\prime}=7.2 N$