An electron and a proton are each placed at rest in an electric ﬁeld of 520N/C. Calculate the speed of each particle 48ns after being released?

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An electron and a proton are each placed at rest in an electric ﬁeld of 520N/C. Calculate the speed of each particle 48ns after being released?

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Answer 1 · 2016-02-23T02:32:11

So the electron final speed is $V_{e} = 4.4 \times 10^{6} \frac{m}{s}$ $V_e = 4.4×10^6 m/s$ and the
proton’s ﬁnal speed is $v_{p} = 2.4 \times 10^{3} \frac{m}{s}$ $v_p = 2.4×10^3 m/s$ ,
$v_{e} \approx 1833 v_{p}$ $v_e ~~ 1833 v_p$

Explanation:

The force on the electron and proton are caused by the E field causing an acceleration that is proportional the mass to respective masses. From $F = q E$ $F = qE$ , and the fact that the magnitude of the electron’s charge is $1.60 \times 10^{- 19} C$ $1.60×10^-19 C$ , the magnitude of the force on the electron is $F = | q | E = (1 .60 \times 10^{- 19} C) (520 \frac{N}{C}) = 8.32 \times 10^{- 17} N$ $F = |q|E = (1 .60×10^-19 C)(520N/C) = 8.32×10^-17 N$
Now the mass of the electron is $m_{e} = 9 .11 \times 10^{- 31} k g$ $m_e =9 .11 × 10^-31 kg$ , from Newton’s 2nd Law, the magnitude of its acceleration is then:
$a_{e} = \frac{F}{m_{e}} = \frac{8.32 \times 10^{- 17} k g \frac{m}{s^{2}}}{9.11 \times 10^{- 31} k g} = 9 .13 \times 10^{13} \frac{m}{s^{2}}$ $a_e = F/m_e =(8.32×10^-17 kg m/s^2)/(9.11×10^-31 kg)=9 .13×10^13 m/s^2$
Now the electron started from rest and reaches a final speed of $v_{e_{f}} = V_{e_{i}} + a t, with V_{e_{i}} = 0$ $v_(e_f) = V_(e_i) + at, " with " V_(e_i) = 0$ we have
$V_{e_{f}} = a t; (9 .13 \times 10^{13} \frac{m}{s^{2}}) (48 \times 10^{- 9} s) = 4.4 \times 10^{6} \frac{m}{s}$ $V_(e_f) = at; (9 .13×10^13 m/s^2) (48×10^-9 s) = 4.4×10^6 m/s$
Similarly for the proton that a mass of $1.67 \times 10^{- 27} k g$ $1.67×10^-27 kg$ , it will end up having an acceleration about 1000 times smaller
$a_{p} = \frac{F}{m_{p}} = \frac{8.32 \times 10^{- 17} k g \frac{m}{s^{2}}}{1.67 \times 10^{- 27} k g} = 4 .98 \times 10^{10} \frac{m}{s^{2}}$ $a_p = F/m_p = (8.32×10^-17 kgm/s^2)/(1.67×10^-27 kg) =4 .98×10^10 m/s^2$
And then the magnitude of the velocity 48ns after being released is
$v = a_{p} t = (4 .98 \times 10^{10} \frac{m}{s} 2) (48 \times 10^{- 9} s) = 2.4 \times 10^{3} \frac{m}{s}$ $v = a_pt = (4 .98×10^10 m/s2)(48×10^-9 s) = 2.4×10^3 m/s$ . ABout

So the electron final speed is $V_{e} = 4.4 \times 10^{6} \frac{m}{s}$ $V_e = 4.4×10^6 m/s$ and the
proton’s ﬁnal speed is $v_{p} = 2.4 \times 10^{3} \frac{m}{s}$ $v_p = 2.4×10^3 m/s$ ,
$v_{e} \approx 1833 v_{p}$ $v_e ~~ 1833 v_p$

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An electron and a proton are each placed at rest in an electric ﬁeld of 520N/C. Calculate the speed of each particle 48ns after being released?

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