Question

# Two point charges $$2q$$ and $$8q$$ are placed at a distance $$r$$ apart. Where should a third charge $$-q$$ be placed between them, so that the electrical potential energy of the system is minimum?

A
At a distance of r/3 from 2q
B
At a distance of 2r/3 from 2q
C
At a distance of r/16 from 2q
D
None of the above

Solution

## The correct option is C None of the aboveLet the third charge $$-q$$ is placed at a distance of $$x$$ from the charge $$2q$$. Then, the potential energy of the system is$$U=k\left[ \dfrac { 2q\left( 8q \right) }{ r } -\dfrac { \left( 2q \right) \left( q \right) }{ x } -\dfrac { \left( 8q \right) \left( q \right) }{ r-x } \right]$$Here, $$k=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } }$$$$U$$ is minimum, where $$\dfrac { 2 }{ x } +\dfrac { 8 }{ r-x }$$ is maximumLet $$\dfrac { 2 }{ x } +\dfrac { 8 }{ r-x } =y$$For $$y$$ to be maximum, $$\dfrac { dy }{ dx } =0$$$$-\dfrac { 2 }{ { x }^{ 2 } } +\dfrac { 8 }{ { \left( r-x \right) }^{ 2 } } =0$$$$\Rightarrow \dfrac { x }{ r-x } =\sqrt { \dfrac { 2 }{ 8 } } =\dfrac { 1 }{ 2 } \Rightarrow x=\dfrac { r }{ 3 }$$But at $$x=\dfrac { r }{ 3 } ,\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =$$ positive i.e. at $$x=\dfrac { r }{ 3 }$$, $$y$$ is minimum or $$U$$ is maximum. So, there cannot be any point between them where the potential energy is minimum.Physics

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