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# An Accelerating Positive Charge: Two Models?

A proton moves at 3.65 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 6.15 103 N/C. (Ignore any gravitational effects in your calculations.)

(a) Find the time interval required for the proton to travel 5.05 cm horizontally.

Δt = _

Since there is no acceleration in the horizontal direction, the horizontal speed must remain constant. How is speed defined?

(b) Find its vertical displacement during the time interval in which it travels 5.05 cm horizontally.

h = _m

Start by determining the vertical acceleration of the proton due to the force it experiences when it enters the electric field. From there you can apply a kinematic expression. Note: You'll need the mass and charge of a proton.

(c) Find the vertical component of its velocity after it has traveled 5.05 cm horizontally.

vy =_m/s

Start by determining the vertical acceleration of the proton due to the force it experiences when it enters the electric field. From there you can apply a kinematic expression. Note: You'll need the mass and charge of a proton.

(d) Find the horizontal component of its velocity after it has traveled 5.05 cm horizontally.

vx =_m/s

### 1 Answer

- Anonymous2 months ago
Welcome to Yahoo!Answers.

(a) t = d / v = 0.0505m / 3.65e5m/s = 1.38e-7 s

(b) acceleration a = qE / m

a = 1.6e-19C * 6.15e3N/C / 1.67e-27kg

a = 5.89e11 m/s²

so

s = ½at² = ½ * 5.89e11m/s² * (1.38e-7s)²

s = 5.64e-3 m = 0.564 cm

(c) v = at

v = 5.89e11m/s² * 1.38e-7s = 8.15e4 m/s

(d) unchanged at 3.65e5 m/s

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