A proton is acted on by an uniform electric field of magnitude 443 N/C pointing

Question

A proton is acted on by an uniform electric field of magnitude 443 N/C pointing in the negative x direction. The particle is initially at rest.

(a) In what direction will the charge move?
---Select--- +x direction ?x direction +y direction ?y direction +z direction ?z direction

(b) Determine the work done by the electric field when the particle has moved through a distance of 3.15 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). F=qE

q is positive here because proton is positively charged. E is in -ve X direction.

Ans: the charge will move in -ve X direction.

B). W=q.E.distance

=1.6*10-19*443*3.15*10-2 (in joule) = 2.232*10-18 joules

Work done by electric field is positive here.

C). U=q.E.d= -2.232*10-18 joules

D). d= 1/2 * (qE/m) * t2

v=0+ (qE/m) * t

mass of proton=  1.673 x 10^?27 kg

So, v= 5.17 * 104 m/s in -ve X direction.

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