Grade 12Electrostatics

Two points charges placed at a distance r in air exert a force f the distance at which they exert Same force when placed in a certain medium

RANJEET Choudhary

8 Years agoGrade 12

2 Answers

Arun

8 Years ago

Dear Ranjeet

F (air) = (1/4 pi eo) [qQ/r^2]
F(medium) = (1/4 pi eo*k) [qQ/R^2]
given> F(medium) = F(air)
(1/4 pi eo*k) [qQ/R^2] = (1/4 pi eo) [qQ/r^2]
(1/k) [1/R^2] = [1/r^2]
kR^2 = r^2
R^2 = r^2/k
R = r/ [rootk]

Regards

Arun (askIITians forum expert)

Rishi Sharma

6 Years ago

Dear Student,
Please find below the solution to your problem.

F (air) = (1/4 pi eo) [qQ/r^2]
F(medium) = (1/4 pi eo*k) [qQ/R^2]
given
F(medium) = F(air)
(1/4 pi eo*k) [qQ/R^2] = (1/4 pi eo) [qQ/r^2]
(1/k) [1/R^2] = [1/r^2]
kR^2 = r^2
R^2 = r^2/k
R = r/ [rootk]

Thanks and Regards