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.

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

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation: