Question

two particle having equal masses of 5.0g each and opposite charge of + 4.0 × 10^-5 coulomb and -4.0×0^-5 c they are released from rest with the separation of 1.0 M between them find the speed of the particle with the separation is reduced to 50 cm

Answer 1

Two particles having equal masses and opposite nature charges are sperated 1 m from each other. both experience electrostatic force of attraction, due to this they will approach.
Let the speed of particle when the separation reduced to 50cm.

Here it is clear that there is no external force so, we have to use conservation of energy theorem.
Change in kinetic energy + change in potential energy = 0
⇒K.Ef - K.Ei = PEi - P.Ef
⇒ 1/2mv² + 1/2 mv² - 0 - 0 = K(q)(-q)/1 - K(q)(-q)/0.5
Here, m = 5g = 0.005kg , q = 4 × 10⁻⁵C
⇒ 0.005 × v² = k(4 × 10⁻⁵)²[ 1/0.5 - 1/1 ]
⇒0.005v² = 9 × 10⁹ × 16 × 10⁻¹⁰/1
⇒ 0.005v² = 14.4
⇒v² = 14.4 × 1000/5
⇒v² = 144 × 100/5
Taking square root both sides,
v = 120/√5 m/s

Answer 2

Hello Dear.

Here is the answer---

We know unlike charges attract each other. Thus, both particles will attract each other.
Given Conditions ⇒

Mass (m) = 5 g.
= 0.005 kg.

Charge (q) = 4 × 10⁻⁵ C.

Original distance between the particles(d₁)= 1 m.
Final distance(d₂) = 50 cm.
= 0.5 m.

Sonce there is no external force we have to use the law of conservation of Energy Theorem.
Thus, change in kinetic energy + change in potential emergy = 0
⇒ KEf - KEi + PEf - PEi = 0.
⇒ KEf - KEi = PEi - PEf.
⇒(1/2 mv² + 1/2 mv²) - (0) = [K (q)(-q)]/d₁ - [K(q)(-q)]/d₂
⇒mv² = [Kq²]/0.5 - [Kq²]/1
⇒ mv² = Kq²[1/0.5 - 1/1]
⇒ 0.005v² = (9 × 10)⁹ × (16 × 10⁻¹⁰)/1
⇒ 0.005 v² = 144/10
⇒ v² = 14.4/(0.005)
⇒ v² = 2880
Takinh square root both sides,
⇒ v = √2880
⇒ v = 24√5
⇒ v = 53.66 m/s.

∴ The speed of the particle is 53.66 m/s.


Hope it helps.