Question

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 1

Answer: 0.0357cm

Step-by-step 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