Question

Two charges Q1 and Q2 are r1 distance apart placed on a smooth horizontal surface .

Charge Q1 is fixed and if Q2 is released from rest then find speed when Q2 is r2 distance away.


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Answer 1

Basic Concept Used :-)

$\sf CONSERVATION \: OF \: MACHENICAL \: ENERGY$

Acc to it ,

$TE_i=TE_f$

$PE_i + KE_i=PE_f+KE_f$

HERE,

Applying conservation of mechanical energy,

$PE_i + KE_i=PE_f+KE_f\\ \\ \implies \frac{kq_1q_2}{ {r _1}^{} } + (0 + 0) = \frac{kq_1q_2}{ {r _2}^{} } + (0 + \frac{1}{2} mv {}^{2} ) \\ \\ \implies \boxed{\boxed{ \sf v = \sqrt{ \frac{2kq_1q_2}{ m }( \frac{1}{r_1} - \frac{1}{r_2} ) } }}$

Which is the required Answer .

Answer 2

Solution :-

At first

We know that

$\sf KE + PE = KE'+ PE'$

Where

KE = Kinetic Energy (Initial)

PE = Potential Energy(Initial)

KE' = Kinetic Energy (Final)

PE' = Potential Energy(Final)

Now

According to the question

KE  = 1/2 mv²

$\sf \dfrac{kQ1 \times Q2}{R1} + 0=\dfrac{kQ1 \times Q2}{R2} + \dfrac{1}{2} mv^2$

$\sf \dfrac{kQ1 + Q2}{R1} + \dfrac{kQ1 + Q2}{R2} = \dfrac{1}{2} mv^2$

$\sf \dfrac{2kQ1Q2}{r1 \times r2} = \dfrac{1}{2}mv^2$

$\sf v = \sqrt{\dfrac{2kQ1Q2}{m} \times \dfrac{1}{R1}-\dfrac{1}{R2}}$