Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conductor, with respect to the outer conductor, is +600 V. An electron is released from rest at the surface of the outer conductor. What is the speed of the electron as it reaches the inner conductor?

Respuesta :

Answer:

v = 1.45 × 10⁷ m/s

Step-by-step explanation:

Given:

Inner radius of the cylinder, r₁ = 20 mm = 0.2 m

outer  radius of the cylinder, r₂ = 80 mm = 0.8 m

Potential difference, ΔV = 600V

Now, the work done (W) in bringing the charge in to the inner conductor

W = [tex]\frac{1}{2}mv^2[/tex]

where, m is the mass of the electron = 9.1 × 10⁻³¹ kg

v is the velocity of the electron

also,

W = qΔV

where,

q is the charge of the electron = 1.6 × 10⁻¹⁹ C

equating the values of work done and substituting the respective values

we get,

qΔV =  [tex]\frac{1}{2}mv^2[/tex]

or

1.6 × 10⁻¹⁹ × 600 = [tex]\frac{1}{2}\times 9.1\times 10^{-31}v^2[/tex]

or

[tex]v = \sqrt\frac{2\times 600\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}[/tex]

or

v = 14525460.78 m/s

or

v = 1.45 × 10⁷ m/s