Question

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

Answer 1

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հίί ʍαtε_______✯◡✯
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һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________
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✯▶from the coloumb's law "for the pair of charges the force of interaction between them is directly proportional to the product of the charges .
____☆now when the charges are altered the charges become 0.9 q and 1.1 q .

 therefore,

♦ new force of interaction between the new pair of charges becomes  F2

▶​writing the above statement in proportion:
"q" is the magnitude
 we get,

 ♦hence F1/ F2 = ( q x q )/ (1.1q x 0.9q)

 or, F1/ F2 = 1/(1.1 x 0.9)

 F1/ F2 = 1/0.99 = 1.01

 so F2 = F1 / 1.01

 ▶given F1 = 100 N

 ♦so F2 = (100/ 1.01) N

 or F2 = 99 N
 
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Answer 2

▶from the coloumb's law "for the pair of charges the force of interaction between them is directly proportional to the product of the charges .

____☆now when the charges are altered the charges become 0.9 q and 1.1 q .

therefore,

♦ new force of interaction between the new pair of charges becomes F2

▶writing the above statement in proportion:

"q" is the magnitude

we get,

♦hence F1/ F2 = ( q x q )/ (1.1q x 0.9q)

or, F1/ F2 = 1/(1.1 x 0.9)

F1/ F2 = 1/0.99 = 1.01

so F2 = F1 / 1.01

▶given F1 = 100 N

♦so F2 = (100/ 1.01) N

or F2 = 99 N