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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
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B
rk
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C
rk
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D
none of these
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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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