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?

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:

We know from accelerated motion that

In our case the proton is released from rest, so and we get

Substituting, we get our final velocity equation:

For the proton we know that and . Writing in S.I. d=0.0025m, we obtain:

frequency or wavelenth of em radiation in a vacuum

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