A proton is launched from an infinite plane of charge with surface charge density -1.90×10-6 C/m2....

A proton is launched from an infinite plane of charge with surface charge density -1.90×10^-6 C/m². If the proton has an initial speed of 3.70×10⁷ m/s, how far does it travel before reaching its turning point?

1.20×10^-6 m

1.80×10^-6 m

66.6 m

133 m

Solutions

Expert Solution

The electric field of infinite plane of surface charge density σ is

in region x>0, where the negative sign indicate that the direction in the negetive x-direction.

Force on charge q in presence of electric field E is qE. The force on the proton is

By Newton's second law of motion, the acceleration of the proton is

The acceleration of the proton is constant. the motion of proton is governed by the kinematics equations for uniformly accelerated motion.

Initial velocity of the proton is , at the turning point the velocity of the proton is zero, v=0.

Solving for the distance travelled

Substituting , we get

genius_generous answered 1 year ago

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