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Question: Two point charges placed at a distance r in air exert force F on each other. The value of distance R...

Two point charges placed at a distance r in air exert force F on each other. The value of distance R at which they experience force 4F when placed in a medium of dielectric constant K = 16 is

A

r

B

r/4

C

r/8

D

2r

Answer

r/8

Explanation

Solution

F = also, 4F = q1q24πε0(16)R2\frac { \mathrm { q } _ { 1 } \mathrm { q } _ { 2 } } { 4 \pi \varepsilon _ { 0 } ( 16 ) \mathrm { R } ^ { 2 } }

\ 14\frac { 1 } { 4 }= 16R2r2\frac { 16 R ^ { 2 } } { r ^ { 2 } } \ R = r8\frac { \mathrm { r } } { 8 }