Question

Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1 µC, m = 47 mg, r1 = 0.83 mm, and r2 = 2.5 mm.

_________m/s

Answer 1

`v_(f)=1721.1m/s`

Step-by-step explanation:

Given data

q = 3.1 µC

m = 47 mg

r1 = 0.83 mm

r2 = 2.5 mm.

As we know that:

`dK=-dU\\(1/2)mv_(f)^(2)-(1/2)mv_(i)^(2)=-((kq^(2) )/(r_(f))-(kq^(2) )/(r_(i)) )\\ (1/2)*(47*10^(-6)kg )v_(f)^(2)-(1/2)*(47*10^(-6)kg)(0)=-[(((9*10^(9)(3.1*10^(-6))^(2) )/(2.5*10^(-3))-(((9*10^(9)(3.1*10^(-6))^(2) )/(0.83*10^(-3)) )]\\2.35*10^(-5) v_(f)^(2)=69.61\\v_(f)=\sqrt{(69.61)/(2.35*10^(-5)) }\\v_(f)=1721.1m/s`