0

Guest

Courses New

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

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

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
 For a pair of charges the force of interaction between them is directly proportional to the product of the charges if the distance is constant.( From coulombs' law)
 
that is hence if initially the charges have a magnitude q
 
and the force of interaction between them is F1
 
now when the charges are altered the charges become 0.9 q and 1.1 q  so the new force of interaction between the new pair of charges becomes  F2
 
​writing the above statement in proportion:
 
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
 
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

Think You Can Provide A Better Answer ?

Other Related Questions on Electrostatics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More