Question

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?

Answer 1

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 = $\frac{1}{2}mv^2$

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 =  $\frac{1}{2}mv^2$

or

1.6 × 10⁻¹⁹ × 600 = $\frac{1}{2}\times 9.1\times 10^{-31}v^2$

or

$v = \sqrt\frac{2\times 600\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}$

or

v = 14525460.78 m/s

or

v = 1.45 × 10⁷ m/s