Question

The distance between two point charges is increased by 10% the force of interaction

Answer 1

Answer:

Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value. If the charges come 10 times closer, the size of the force increases by a factor of 100. The size of the force is proportional to the value of each charge.

Answer 2

The force of interaction will decrease by 17.35 % .

Let two point charges be Q and q .

Let initial distance between then is r .

Therefore , Force between the charges be :

$F=\dfrac{kqQ}{r^2}$

Here , k is constant .

Now , when distance between them is increased by 10 % .

Therefore , new length $r'$ is :

$r'=r+\dfrac{10}{100}r\\\\r'=\dfrac{11}{10}r$

Therefore , new force $F'$ be :

$F'=\dfrac{kqQ}{r'^2}\\\\F'=\dfrac{kqQ}{\dfrac{11^2 r^2}{10^2}}\\\\\\F'=\dfrac{100}{121}\times \dfrac{kqQ}{r^2}$

Therefore , % decrease :

$\dfrac{F-F'}{F}\times 100\\\\\dfrac{\dfrac{kqQ}{r^2}-\dfrac{100}{121} \dfrac{kqQ}{r^2}}{\dfrac{kqQ}{r^2}}\times 100\\\\\\\dfrac{21}{121}\times 100\\\\17.35\ \%$

Hence , this is the required solution .

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Laws Of Motion

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