Question

TheE-field strength is 50,000 N/C inside a parallel plate capacitor with 2.0 mmspacing. A proton is released from rest at the positive plate. What is the proton’sspeed when it reaches the negative plate?

Answer 1

Answer:

The speed is $138852.4 \frac{m}{s}$

Explanation:

Electric field magnitude (E) and electric force magnitude (F) are related by the equation

$F=Eq$

with q the charge of the proton ($1.61\times10^{-19} C$). Because between parallel plates electric field is almost constant, electric force is constant too:

$F=(50000)(1.61\times10^{-19})=8.05\times10^{-15}N$

because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:

$a=\frac{F}{m}$

with m the mass of the proton ($1.67\times10^{-27} kg$):

$a=\frac{8.05\times10^{-15}}{1.67\times10^{-27}}= 4.82\times10^{12}\frac{m}{s^{2}}$

Now with this constant acceleration we can use the kinematic equation

$v^{2}=v_{0}^{2}+2ad$

with v the final speed, $v_i$ the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:

$v=\sqrt{2ad}=\sqrt{2(4.82\times10^{12}\frac{m}{s^{2}})(2.0\times10^{-3}m)}$

$v=138852.4 \frac{m}{s}$