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

Answer:

a) 800 keV

b) 25 km

Explanation:

$Strength\ of\ Electric\ field=2\times 10^6\ V/m\\a)\ Potential\ Difference=Strength\ of\ Electric\ field\times Distance\\\Rightarrow Potential\ Difference=Kinetic\ Energy\ =2\times 10^6\times 0.4\\\therefore Energy=0.8\times 10^6\ eV=800\ keV$

$b)\ Potential\ difference=50\ GeV=50\times 10^9\ eV\\Distance=\frac{Potential\ difference}{electric\ field}\\\Rightarrow Distance=\frac{50\times 10^9}{2\times 10^6}\\\Rightarrow Distance=25\times 10^3\ m\\\therefore Distance=25\ km$

Answer 2

Answer:

a) 800 keV

b)  24.996 km.

Explanation:

(a) we have

$\large \Delta K.E=q\Delta V$  .............(1)

where,

$\large \Delta K.E$ = Change in kinetic energy

$q$ = charge of an electron

$\Delta V$ = Potential difference

also

$\large E=\frac{V}{d}$       .......(2)

E = electric field

d = distance traveled

Now from (1) and (2) we have,

$\large \Delta K. E=qV=qEd$

substituting the values in the above equation, we get

$\large \Delta K. E=(1.6\times 10^{-19}C)(2\times 10^6V/m)(0.400m)(\frac{1eV}{1.6\times 10^{-19}J})(\frac{1keV}{1000eV})$

$\large \Delta K. E=800keV$

Thus, the energy gained by the electron is 800 keV if it is accelerated over a distance of 0.400 m.

(b) Using the equation (1), we have

$\large d=\frac{\Delta K.E}{qE}$

$\large d=\frac{(50\times 10^9eV)}{(1.6\times 10^{-19C})(2\times 10^6V/m)}(\frac{1.6\times 10^{-19}J}{1eV})$

or

$\large d=2.4996\times 10^4m$

or

$\large d=24.996\times 10^3m=24.996km$

Thus, to gain 50.0 GeV of energy the electron must be accelerated over a distance of 24.996 km.