Question

The electric field strength is 5.50×10^4 N/C inside a parallel-plate capacitor with a 2.50 mm spacing. A proton is released from rest at the positive plate.What is the proton's speed when it reaches the negative plate?

Answer 1

Answer:

$v=1.6\times10^{5}m/s$

Explanation:

The equation that relates electric field strength and force that a charge experiments by it is F=qE.

Newton's 2nd Law states F=ma, which for our case will mean:

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

We know from accelerated motion that $v^2=v_0^2+2ad$

In our case the proton is released from rest, so $v_0=0m/s$ and we get $v=\sqrt{2ad}$

Substituting, we get our final velocity equation:

$v=\sqrt{\frac{2qEd}{m}}$

For the proton we know that $q=1.6\times10^{-19}C$ and $m=1.67\times10^{-27}kg$. Writing in S.I. d=0.0025m, we obtain:

$v=\sqrt{\frac{2(1.6\times10^{-19}C)(5.5\times10^{4}N/C)(0.0025m)}{(1.67\times10^{-27}kg)}}=162318.5m/s=1.6\times10^{5}m/s$