These two are giving me a hard time A proton is released from rest In a uniform

Question

These two are giving me a hard time A proton is released from rest In a uniform electric field of magnitude 409 N/C. (a) Find the electric force on the proton. (b) Find the acceleration of the proton. (c) Find the distance it travels in 1.88 mus. A point charge q = +36.0 muC moves from A to B separated by a distance d = 0.163 m in the presence of an external electric field E of magnitude 270 N/C directed toward the right as in the following figure. (a) Find the electric force exerted on the charge. (b) Find the work done by the electric force. (c) Find the change in the electric potential energy of the charge. (d) Find the potential difference between A and B. VB - VA = V

Explanation / Answer

Here ,

a) electric field , E = 409 N/C

force on the proton , F = e * E

F = 1.602 *10^-19 * 409

F = 6.553 *10^-17 N

the electric force on the proton is 6.553 *10^-17 N

b)

using second law of motion

a = F/m

a = 6.553 *10^-17 /(1.67 *10^-27)

a = 3.92 *10^10 m/s^2

the acceleration of proton is 3.92 *10^10 m/s^2

c)

using seocond equation of motion

d = 0.5 at^2

d = 0.5 * 3.92 *10^10 *(1.88*10^-6)^2

d = 0.069 m = 6.9 cm

distance travelled by proton is 6.9 cm

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