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Two point charges exert on each other a force F when they are placed r distance apart in air. When they are placed R distance apart in medium of dielectric constant K theyexert the same force . The distance R equals

Two point charges exert on each other a force F when they are placed r distance apart in air. When they are placed R distance apart in medium of dielectric constant K theyexert the same force . The distance R equals
 

Grade:11

3 Answers

Arun
25750 Points
6 years ago
Dear student
1/4 π ε0 q1* q2/ r2 = 1/4 π ε0 q1*q2 / K*R2
1/r2 = 1/ K*R2
R = r √k
Regards
Arun (askIITians forum expert)

π ε0

laksanya
13 Points
5 years ago
F=kq1q2/r^2
so if r=>r/2
let the new force be F'
F' = kq1q2/(r/2)^2
F'=4kq1q2/r^2
F'=4F

so force changes to 4F
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

1/4 π ε0 q1* q2/ r2 = 1/4 π ε0 q1*q2 / K*R2
1/r2 = 1/ K*R2
R = r √k

Thanks and Regards

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