Question

two point charges repel each other with a force of 100N .if one charge is increased by 10% and other is reduced by 10% .the new force of repulsion at the same distance would be..?​

Answer 1

Answer:

Question:-

      ·  Two point charges repel each other with a force of 100 N .if one charge is increased by 10% and other is reduced by 10% .the new force of repulsion at the same distance would be?​

Solution:-

100 = $\frac{K q_{1} q_{2} }{r^{2} }$

Now, $q_{1}$ ⇒ 0.9

        $q_{2}$  ⇒ 1.1  

$F^{1} = \frac{K (0.9 q_{1})(1.1 q_{2} )}{r^{2} }$

$\frac{F^{1} }{100} = \frac{0.9 X 1.1 }{1}$  ⇒  $F^{1} = \frac{0.9}{10}$ × $\frac{1.1}{10}$ × 100 = 99

$F^{1}$ = 99 N

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