An electric field of strength E = 6900 N/C Is directed along the +x-axis as show

Question

An electric field of strength E = 6900 N/C Is directed along the +x-axis as shown. A proton is initially at rest at point A. What is the magnitude of the force on the proton due to the electric field? 1.106 times 10^-15 N What is the magnitude of the acceleration of the proton? 660939608590.5840 m/s^2 What is the work done by the electric field on the proton after it has moved a distance of 6 cm from point A to point B? 6.633 times 10^-17 What is the kinetic energy of the proton at this instant? _________ J What is the speed (magnitude of the velocity) of the proton at this instant? ________ m/s

Explanation / Answer

work done by the electric field on the proton after it has moved a distance of 6 cm from pointA to point B is given as   

   W = 6.633*10^-17 J

this is equal to change in kinetic energy of the proton so, K.E = 0.5*mv^2 = 6.633*10^-17 J

now the speed of the protonn is

   V = sqrt(2K.E/m)

   V = sqrt(2*6.633*10^-17/(1.6726219*10^-27)) m/s

   V = 281624.94 m/s

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