Question

A proton moves at 4.30 x 10 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 7.90 10 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 4.70 cm horizontally. (b) Find its vertical displacement during the time interval in which it travels 4.70 cm horizontally (c) Find the horizontal and vertical components of its velocity after it has traveled 4.70 cm horizontally Part 1 of 7-Conceptualize The proton maintains a constant velocity in the horizontal direction. In less than a microsecond, it could traverse a vacuum tube while moving with a huge vertical acceleration in order to experience a measurable deflection. Part 2 of 7-Categorize This is a projectile motion problem. One component of the motion has zero acceleration and the other component has constant nonzero acceleration. Part 3 of 7-Analyze The electric field i is directed along the y axis; therefore, we know that o and (a) Solving for t, we have the following 0.047m 10 ms (b) We apply Newton's second law in the vertical direction to the electric force component o, to obtain qe, kg

Answer 1

(a) Since the force on the proton due to the electric field will be in the vertical direction, there will be no change in the velocity component along the horizontal direction. Therefore the time taken by the proton to travel a distance of 4.70 cm is,

$t = \frac{distance}{speed} = \frac{4.70 \times 10^{-2}}{4.30 \times 10^5} = 1.09 \times 10^{-7} \,s = 0.109 \mu s$

(b) The force on the proton along the vertical direction is,

$F = qE \Rightarrow F = (1.6 \times 10^{-19})(7.90 \times 10^3) = 1.264 \times 10^{-15} \, N$

Therefore the acceleration of the proton along that direction is,

$\Rightarrow a = \frac{F}{m_p} = \frac{1.264 \times 10^{-15}}{1.67 \times 10^{-27}} = 7.57 \times 10^{11} \, m/s^2$

Now the vertical displacement in 0.109 microseconds is,

$s = ut + \frac{1}{2}at^2 \Rightarrow s = \frac{1}{2}(7.57 \times 10^{11})(0.109 \times 10^{-6})^2 = 4.5 \times 10^{-3} \, m = 45 \,cm$

(c) There will be no change in the velocity along the horizontal direction so after the proton has traveled 4.70 cm, the horizontal velocity will remain the same.

The vertical velocity of the proton after it has traveled 4.70 cm in the horizontal direction (ie. time taken is 0.109 microseconds) is given by,

$v = u + at \Rightarrow v = at \Rightarrow v = (7.57 \times 10^{11})(0.109 \times 10^{-6}) = 8.25 \times 10^{4} \, m/s$