Protons are released from rest at an electric potential of 5 MeV obtained from a

Question

Protons are released from rest at an electric potential of 5 MeV obtained from a Van de Graaff accelerator and travel through a vacuum to a region to a region at zero potential. If this change in potential occurs over a distance of 2.0 m, find the electric field, assuming it to be uniform. Find the speed of the 5-MeV protons. (The proton mass is 1.67 times 10-27 kg.) The potential due to a charge Q uniformly distributed on a spherical conductor of radius R is kQ/r for What is the potential in the region A spherical conductor has a radius r = 10 cm and carries charge of 0.2 nC. Draw to scale the equipotential surfaces in potential steps of 2 V.

Explanation / Answer

Given Initial velocity of proton is , vi = 0 m/s Initial potential energy , Ui = 5 *10^6 eV Since 1eV = 1.6 *10^-19 J Thus, Ui = ( 5 *10^6 eV) ( 1.6 *10^-19 J) /1eV = 8 *10^-13 J Change in potential energy is, U = 8 *10^-13 J This change occurs over a distance of , d = 2.0 m a) From the relation     U = V q            = E d q     E = U / d*q = 8 *10^-13 J / 2.0 m * 1.6 *10^-19 C     E = 2.5 *10^6 N/ C The magnitude of uniform electric field is 2.5 *10^6 N/ C ________________________________________________ b) By law of conservation of energy   Change in potential energy = kinetic energy           U = 1/2 mv^2 where m is the mass of proton , m = 1.67 *10^-27 kg           v^2 =   2U / m                 = 2 * (8 *10^-13 J) /1.67 *10^-27 kg            v ˜ 3.0 *10^7 m/s The speed of the proton is   3.0 *10^7 m/s               

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