Physics, 14.09.2019 06:30 Yellowjacket3166

Calculate the speed of a proton that has moved in a uniform electric field of 180.0 n/c from rest over a distance of 12.5-cm. assume it began at rest.

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Answer from: rajenkins79

Given:

Electric field = 180 N/C

$Force\ on\ proton = 1.6\times10^{-19} C$

$Force\ on\ proton = 180\times1.6\times10^{-19} =288\times10^{-19} N$

$Mass\ of\ proton = 1.673\times10^{-27} kg$

$Acceleration of proton = \frac{force}{mass}$

$Acceleration\ of\ proton = \frac{288\times10^{-19}}{1.673*10^{-27}} =172\times108 m/s^{2}$

Let the speed of proton be "x"

x = $\sqrt{Acceleration}$

$x = \sqrt{(2\times172\times108\times0.125)}=65602.2 m/s$

Answer from: justalikri

the velocity of the proton is 65574.38 m/s

Explanation:

given,

uniform electric field = 180 N/C

Distance = 12.5 cm = 0.125 m

charge of proton = 1.6 × 10⁻¹⁹ C

force = E × q

=180 × 1.6 × 10⁻¹⁹

F= 2.88 × 10⁻¹⁷ N

mass of proton = 1.673 × 10⁻²⁷ kg

acceleration = $\dfrac{force}{mass}$

= $\dfrac{2.88 \times 10^{-17}}{1.673\times 10^{-27}}$

=1.72 × 10¹⁰ m/s²

velocity = $\sqrt{2\times 0.125 \times 1.72 \times 10^{10}}$

=65574.38 m/s

hence , the velocity of the proton is 65574.38 m/s

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