Q. 694.6( 5 Votes )

Two particles have equal masses of 5.0g each and opposite charges of +4.0 × 10–5 C and –4.0 × 10–5 C. They are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.

Answer :


Given:
Mass of the two particles : m1=m2= m = 5.0 g = 0.005 kg
Charge on particle 1 : q1 = +4.0 × 10–5 C
Charge on particle 2 : q2 = -4.0 × 10–5 C
Separation between the charges : r = 1 m
Initial velocity: v = 0
Formula used:
Conservation of Energy is:
Uinitial + K.Einitial = Ufinal + K.Efinal
where, Uinitial and K.Einitial are the Initial potential energy and initial kinetic energy respectively .
Ufinal and K.E final are the final potential energy and final kinetic energy respectively.
Now, as initial velocity is zero : K.Einitial = 0
Final Kinetic Energy of both the particles is:
Also, Electric potential energy is given as;
Where k is a constant and k= =9× 109 Nm2C-2 , q1 and q2 are point charges and r is the separation between them.
For initial situation : r = 1 m
For final situation : r = 50 cm = 0.5 m = r/2
Substituting in the conservation formula we get,






Hence the velocity of the particles when separation is reduced to 50 cm is 53.66 m/s


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