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2 years ago

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A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next

to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed

1 answer:

zalisa [80]2 years ago

5 0

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

$V=1.4*10^5m/s$

Explanation:

From the question we are told that:

Electric field $B=1.5*10N/C$

Distance $d=2 x 10^{-3}$

At negative plate

Generally the equation for Velocity is mathematically given by

$V^2=2as$

Therefore

$V^2=\frac{2*e_0E*d}{m}$

$V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}$

$V=\sqrt{19.2*10^9}$

$V=1.4*10^5m/s$

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