Two spherical conductors, A and B , are placed in vacuum. A has a radius rA=25cm and B of rB=35cm. The distance between the centers of the two spheres is d=225 cm.

A has a potential of VA =120 Volt and B has a potential of VB = -40 Volt.

a. An electron is released with zero speed from A . What will its speed be as it reaches B ? Express your answer in m/sec.

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b. A proton is released with zero speed from A . What will its speed be as it reaches B ? Express your answer in m/sec.

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c. We now change the potential of B to VB=+25 Volt. For this new configuration, What is the ratio of the speed of the electron (as it arrives at A ) and the speed of the proton (as it arrives at B )?

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anyone?

Please some one answer.

To answer these questions, we will need to use the principles of electric potential and conservation of energy. The potential difference between two points is defined as the work done per unit charge in moving a positive test charge from one point to another.

a. To find the speed of an electron as it reaches B, we can use the principle of conservation of energy. The initial potential energy of the electron at A is given by qVA, where q is the charge of an electron and VA is the potential at A. The final kinetic energy of the electron at B is given by (1/2)mv^2, where m is the mass of an electron and v is its final speed. Since the potential energy is converted into kinetic energy, we can equate these two expressions:

qVA = (1/2)mv^2

To find the value of v in m/s, we need the value of q (charge of an electron) and m (mass of an electron). The charge of an electron is approximately 1.602 x 10^-19 C, and the mass of an electron is approximately 9.109 x 10^-31 kg.

b. Similar to part a, we can use the principle of conservation of energy to find the speed of a proton as it reaches B. The initial potential energy of the proton at A is given by qVA, where q is the charge of a proton (equal to the charge of an electron but with a positive sign) and VA is the potential at A. The final kinetic energy of the proton at B is given by (1/2)mv^2, where m is the mass of a proton and v is its final speed. Equating these two expressions, we can solve for v.

c. To find the ratio of the speeds of the electron and the proton in the new configuration, we need to find the speed of the electron as it arrives at A and the speed of the proton as it arrives at B. We can use the same principles of conservation of energy to find these speeds, but with a different potential difference. The potential at A remains the same (VA = 120 V), but the potential at B has changed to VB = 25 V.

By solving these equations and plugging in the values, we can find the answers to all the questions.