Question

Two electrons are initially held at rest with a separation of 4.0 μm. If one of the electrons is released, how fast will it be traveling when it reaches a distance of 1.0 mm from the other? Give your answer in m/s to three significant figures.

Answer 1

Sol::

Given data:

Since both elctrons have same (negative) charge,the charges repels each other.

Now using the conservation of energy theorem

We have
Change in kinetic energy + Change in Electrostatic Energy = 0

⇒(Electrostatic energy)f - (electristatic energy)i =(KEi -KEf)

As intake electrons are at rest so Vi=0
i. e KEi=0

Final distance (d₂) = 0.001004 m

Initial distance (d₁) = 0.000004 m

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

∴ kq²/d₂ - kq²/d₁ = 0 - 2 × 0.5× mv²

kq²[1/d₂ - 1/d₁] = - mv²
(9*10^9)(1.60*10^-19)²[1/0.001004-1/0.000004]=
-(9.11*10^-31)v²
v²= 393595.7611
v= 6273.72 m/s

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