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A 2.6 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away comes to within 0.37 mm of the surface of the target before being reflected. What was the electron's initial speed? At what distance from the surface of the sphere is the electron's speed half of its initial value? What is the acceleration of the electron at its turning point?

User Arian
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Final answer:

The problem requires applying the conservation of energy principle to find the initial speed and the distance at which the electron's speed is reduced to half. It also involves using Coulomb's law to calculate the electron's acceleration at the closest approach to the charged sphere.

Step-by-step explanation:

The student is dealing with a physics problem involving electric fields and forces on charged particles. To find the initial speed of an electron approaching a charged sphere, the conservation of energy principle must be utilized. We equate the kinetic energy of the electron far away from the sphere with the potential energy at the closest approach plus remaining kinetic energy, if any. To find where the speed of the electron is half of the initial value, we again apply the conservation of energy, knowing that at that point the kinetic energy will be one-fourth of the initial kinetic energy because kinetic energy is proportional to the square of the speed.

The acceleration of the electron at its turning point can be found by understanding that at the turning point, all kinetic energy has been converted to potential energy, and using the electric force at that position which is given by Coulomb's law.

The essential details required to calculate these variables include the charge on the sphere, the charge and mass of the electron, the initial and turning point distances. Given that the initial potential energy is zero (at an infinite distance), the entire initial kinetic energy of the electron will be converted into electric potential energy at the turning point.

User Gejun
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