A proton is acted on by an uniform electric field of magnitude 193 N/C pointing

Question

A proton is acted on by an uniform electric field of magnitude 193 N/C pointing in the positive x direction. The particle is initially at rest.

(a) In what direction will the charge move?
1---Select---+x directionx direction+y directiony direction+z directionz direction = +x direction

(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.
2. ? J

(c) Determine the change in electric potential energy of the charged particle.
3. ? J

(d) Determine the speed of the charged particle.
4. 37218 m/s

Explanation / Answer

here,

electric feild , E = 193 N/C

(a)

the charge will move in -x direction

(b)

force acting on proton , F = E * e

F = 193 * 1.6 * 10^-19

F = 3.088 * 10^-17 N

displacement of particle , d = 3.75 cm

d = 0.0375 m

the work done by the electric field , W = F * d

W = 0.0375 * 3.088 * 10^-17

the work done by the electric field is 1.16 * 10^-18 J

(c)

the change in electric potential energy of the charged particle is -1.16 * 10^-18 J

(d)

F = 3.088 * 10^-17 N

accelration , a = F/m

a = 3.088 * 10^-17 / (1.67*10^-27)

a = 1.85 * 10^10

initial velocity , u = 0

distance travelled ,s = 0.0375 m

let the speed of particle be v

using third equation of motion

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

v^2 = 2 * 1.85 * 10^10 * 0.0375

v = 37249 m/s

speed of the particle is 37249 m/s

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