Question

# Two point charges placed at a certain distance r in air exert a force F on each other. The distance r' at which these charges will exert the same force in medium if dielectric constant k is given by

A
r
B
rk
C
rk
D
none of these

Solution

## The correct option is C $$\dfrac{r}{ \sqrt{k}}$$Two point charge placed at a certain distance $$=r$$force $$=F$$dielectric constant $$=k$$Let, two point charges $${ q }_{ 1 }$$ and $${ q }_{ 2 }$$ are separated $$r$$ distance from each other.Then, force experienced between then,$$F=\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }_{ 0 }{ r }^{ 2 } }$$Put, $$F=\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }_{ 0 }{ r }^{ 2 } }$$$$\therefore$$   $$\dfrac { 4{ q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }_{ 0 }{ r }^{ 2 } } =\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ k\left( 4\pi { \epsilon }_{ 0 }{ r }^{ 2 } \right) }$$$$\Rightarrow \dfrac { 1 }{ { r }^{ 2 } } =\dfrac { 1 }{ 4{ r }^{ 2 }h }$$$$\Rightarrow \dfrac { 1 }{ { r }^{ 2 } } =\dfrac { 1 }{ { kr }^{ 2 } }$$taking square root both sides $$\dfrac { 1 }{ r } -\dfrac { 1 }{ \sqrt { k } { r }^{ 1 } }$$or,  $${ r }^{ 1 }=\dfrac { r }{ \sqrt { k } }$$Physics

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