Final answer:
The new speed of the charged ball after being accelerated by a 2000 V potential difference is twice its original speed, meaning the correct answer is option B. (2v). This is due to the fact that kinetic energy is quadrupled when the electric potential is quadrupled, resulting in the speed increasing by a factor of two.
Step-by-step explanation:
When a charged particle is accelerated through a potential difference, it gains kinetic energy proportional to the charge of the particle and the magnitude of the potential difference through which it is accelerated. Using the relationship between kinetic energy and potential energy (K.E. = qV), we can say that when the electric potential is quadrupled (from 500 V to 2000 V), the kinetic energy of the particle is also quadrupled. Because kinetic energy is proportional to the square of the velocity (K.E. = 1/2 mv^2), when the kinetic energy becomes four times greater, the speed of the particle increases by a factor of two. Therefore, if the initial speed is v, the new speed when the potential difference is increased to 2000 V will be 2v, making the correct answer option B.