A proton is acted on by an uniform electric field of magnitude 433 N/C pointing

Question

A proton is acted on by an uniform electric field of magnitude 433 N/C pointing in the negative y direction. The particle is initially at rest. (a) In what direction will the charge move? Incorrect: Your answer is incorrect. What is the direction of the force due to the electric field on the charged particle? (b) Determine the work done by the electric field when the particle has moved through a distance of 3.75 cm from its initial position. J (c) Determine the change in electric potential energy of the charged particle. J (d) Determine the speed of the charged particle. m/s

Explanation / Answer

A.

E = -433 N/C j

q = 1.6*10^-19 J

F = q*E = -1.6*10^-19*433

F = -6.93*10^-17 j

Since force is in -ve y direction acceleration will also be in -ve y direction.

So Charge will move in -ve y direction.

B.

Work = F*s

s = 3.75 cm = 0.0375 m

W = 6.93*10^-17*0.0375

W = 2.6*10^-18 J

C.

Change in potential energy = -W

dPE = -2.6*10^-18 J

D.

a = F/m

a = 6.93*10^-17/(1.67*10^-27)

a = 4.15*10^10 m/sec^2

v^2 = u^2 + 2*a*s

u = 0

v = sqrt (2*a*s)

v = sqrt (2*4.15*10^10*0.0375)

v = 55789.78 m/sec

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