Two electron lying 10cm apart are released.what will be speed when they are at 20 cm appart
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Answered by
22
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 - ( 1/2 mv²+ 1/2mv²) [ 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
∴ 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 - ( 1/2 mv²+ 1/2mv²) [ 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
Answered by
0
Let their velocities at 20cm a part
be v. Then
ΔKE
Total
=2[
2
Me
v
2
−
2
m
(0)
2
]=Mev
2
Potential energy 10 cm (U Initial )
=
100
10
−1ke
2
=10ke
2
Potential Energy 20 cm (U final )
=
20/100
ke
2
=5ke
2
ΔU
Total
=U
final
−U
initial
=5ke
2
−10ke
2
=−5ke
2
As electrostatics force is
conservative,
ΔKE
Total
+ΔPE
Total
=0
⇒mev
2
−5ke
2
=0
⇒v=
me
5ke
2
=35.58m/s
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