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Need explanation for: - Electrostatics - JEE Main

Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d< < l) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity \upsilon . Then as a function of distance x between them

  • Option 1)

    \upsilon\: \alpha\: x^{-1/2}

  • Option 2)

    \upsilon\: \alpha\: x^{-1}

  • Option 3)

    \upsilon\: \alpha\: x^{1/2}

  • Option 4)

    \upsilon\: \alpha\: x

 
Answers (3)
133 Views
S Sujal

Option 1 ans

 

 

 

N neha

As we learnt in 

Coulombic force -

F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}

- wherein

K - proportionality Constant 

Q1 and Q2 are two Point charge

 

 T\cos \Theta =mg

T\sin \Theta =f_{e}=\frac{1}{4\pi\varepsilon _{o}}\:\frac{q^{2}}{d^{2}}

tan \Theta =\frac{1}{4\pi\varepsilon _{o}}\:\frac{q^{2}}{d^{2}mg}

\therefore tan \Theta=\frac{x}{2l}

\frac{x}{2l}=\frac{q^{2}}{4\pi\varepsilon _{o}x^{2}mg}

\Rightarrow \frac{x}{2l}\propto \frac{q^{2}}{x^{2}}

\Rightarrow q^{2}\propto x^{3}

\Rightarrow q\propto x^{3/2}

\Rightarrow\frac{dq}{dt}\propto\frac{3}{2}x^{1/2}\frac{dx}{dt} \therefore \left (\frac{dq}{dt}= constant \right )

\therefore \frac{dx}{dt}=v\propto x^{-1/2}

 


Option 1)

\upsilon\: \alpha\: x^{-1/2}

This is correct option

Option 2)

\upsilon\: \alpha\: x^{-1}

This is incorrect option

Option 3)

\upsilon\: \alpha\: x^{1/2}

This is incorrect option

Option 4)

\upsilon\: \alpha\: x

This is incorrect option

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