Answer to Question #153422 in Electric Circuits for Rashinda

Question

An electron is to be accelerated in a uniform electric field having a strength of 2.00 × 10⁶ V/m.

(a) What energy in keV is given to the electron if it is accelerated through 0.400 m?

(b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?

Answer 1

The expression for the potential difference is

$ΔV = -Ed$

E = the electric field strength between the plates

d = he separation between the plates

$(a) \; ΔV = -(2.00 \times 10^6) \times (0.400) \\ = -8.0 \times 10^5 \;V$

The electric energy is expressed as follows:

$U = qΔV$

q = the charge of the electron

$U = (-1.6 \times 10^{-19})(-8.0 \times 10^5) \\ = 1.28 \times 10^{-13} \;J (\frac{1 \; keV}{1.6 \times 10^{-16} \;J}) \\ = 800 \;keV$

The electric energy is 800 keV.

$(b) \; ΔV = \frac{U}{q} \\ d = -\frac{ΔV}{E} \\ d = -\frac{\frac{U}{q}}{E} \\ = -\frac{U}{qE} \\ d = \frac{(50.0 \;GeV)(\frac{10^9 \;eV}{GeV})(-\frac{1.6 \times 10^{-19} \;J}{1 \;eV})}{(2.00 \times 10^6 \;v/m)(1.6 \times 10^{-19} \;C)} \\ = 25 \times 10^3 \;m \\ = 25 \;km$

The distance covered by the electron is 25 km.