Question

two electrons lying 10cm apart are released. what will be their speed when they are 20 cm apart ??

Answer 1

both the charged particles are same nature so, force acts between them is repulsive force. Due to this both are repel each other and they will move away from each other.

∴ change in electrostatic energy + change in kinetic energy = 0 { from conservation energy theorem}
Final Electrostatic energy - initial electrostatic energy = intial Kinetic energy - speed kinetic energy
⇒ Ke²/(20/100) - Ke²/(10/100) = 0 - 2 × 1/2 mv² [ because intially electrons are rest ]
⇒ 9 × 10⁹ × (1.6 × 10⁻¹⁹)²{1/0.2 - 1/0.1 } = -2/2 × 9.1 × 10⁻³¹ v²
⇒9 × 2.56 × 10⁻²⁹ × (-5) = - 9.1 × 10⁻³¹ v²
⇒9 × 2.56 × 5 × 100/9.1 = v²
⇒v = 35.8 m/s

Answer 2

Hello Dear.

Here is the answer---

Since the elctrons have the negative charge. Thus both charges repel each other.

Since there is no external force acting on the electrons, thus using the conservation of energy theorem.
∴ Change in kinetic energy + Change in Electrostatic Energy = 0
⇒ Final Electrostatic energy - Initial electristatic energy = Initial kinetic energy - Final kinetic enrgy

Initial kinetic energy = 0 [Since, the elctrons are initially at rest]

Final distance (d₂) = 20 cm.
= 0.2 m
Initial distance (d₁) = 10 cm
= 0.1 cm.

k = 9 × 10⁹
Mass of the Electrons (m) = 9.1 × 10⁻³¹
Charge on the Electrons = 1.6 × 10⁻¹⁹

∴ kq²/d₂ - kq²/d₁ = 0 - 2 × 1/2 × mv²
⇒ kq²[1/d₂ - 1/d₁] = - mv²
⇒ 9 × 10⁹ × (1.6 × 10⁻¹⁹)²[ 1/0.2 - 1/0.1] = - 9.1 × 10⁻³¹ v²
⇒ (9 × 2.56 × 10⁻²⁹)(-5) = - 9.1 × 10⁻³¹ v²
⇒23.04 × -5 × 100 = - 9.1 v²
⇒ -11520 = -9.1 v²
⇒ v² = 11520/9.1
Taking square root on both sides.
⇒ v = √1265.93
⇒ v = 35.58 m/s.

∴ Speed of the Electrons is 35.58 m/s.


Hope it helps.