A very long nonconducting cylinder of diameter 10.0 cm carries charge distributed uniformly over its surface. Each meter of length carries + 5.50 µC of charge. A proton is released from rest just outside the surface. How far will it be from the surface of the cylinder when its speed has reached 2550 km/s? The mass of the proton is 1.67 * 10-27 kg and e = 1.60 * 10-19 C.

Answer: 0.0357cm

Explanation:

Given the following;

Diameter of cylinder (d) = 10cm

Radius of cylinder (r) = 5cm

charge density (λ) = 5.50 * 10^-6C/m

Proton speed (v) = 2550km/s=2.55*10^6 m/s

Electric potential at a distance x is given by;

V = q λ / 2 π ε_0 ln (r / x)

The potential energy of proton is given by;

P. E = q λ / 2 π ε_0 ln (r / x)

P. E = K. Eq λ / 2 π ε_0 ln (r / x)

Recall Kinetic Energy (K. E) = 0.5 mv^2

[ (0.5mv^2 * (1.6*10^-19) * (5.5*10^-6) / (2 π * 8.85*10^-12 * ln (5*10^-2 m / x)) ]

[0.5 * 1.67*10^-27 * (2.55*10^6) ^2ln (5.0*10^-2 / x)

0.342 (5.0*10^-2 / x)

= 1.40x = 3.57 * 10^-2 m = 0.0357 cm