Question

Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each other. What is the speed of either electron when their separation has increased by a factor of 4.10?

Answer 1

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Step-by-step explanation:

The electric potential energy is given by the relation :

`U = (kq_(1)q_(2) )/(r)`

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

`U_(1) = (ke^(2) )/(r_(1) )`

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

`U_(2) = (ke^(2) )/(r_(2) )=(ke^(2) )/(4.10r_(1) )`

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

`(1)/(2)mv^(2)=(ke^(2) )/(r_(1) )- (ke^(2) )/(4.10r_(1) )`

Here m and v are the mass and final speed of electron respectively.

`v^(2)=(2)/(m) (ke^(2) )/(r_(1) )(1- (1 )/(4.10 ))`

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

`v^(2)=(2)/(9.1*10^(-31) ) (9*10^(9)*(1.6*10^(-19))^(2) )/(4.32*10^(-10) )(1- (1 )/(4.10 ))`

`v^(2)=8.86*10^(11)`

v = 9.41 x 10⁵ m/s