Question

An electron is initially at rest in a uniform electric field having a strength of 2.1 × 10^6 V/m. It is then released and accelerated by the presence of the electric field. Part A) What is the change in the electron’s kinetic energy, in kiloelectron volts, if it travels over a distance of 0.325 m in this field?

Part B) Over how many kilometers would it have to be accelerated in the same electric field to increase its kinetic energy by 65 GeV?
Part C) In what direction did the electron move?

Answer 1

(A). The change in the electron’s kinetic energy is
`6.8*10^(2)\ k eV`.

(B). The distance is 30.95 km.

(C). The direction of electron is opposite direction of the electric field.

Step-by-step explanation:

Given that,

Electric strength
`F=2.1*10^(6)\ V/m`

Distance = 0.325 m

(A). We need to calculate the change in the electron’s kinetic energy

Using formula of work done

Kinetic energy=Work done by electric force

`K.E=F* d`

Put the value into the formula

`K.E=2.1*10^(6)*0.325`

`K.E=682500\ J`

`K.E=6.8*10^(2)\ k eV`

The change in kinetic energy is
`6.8*10^(2)\ k eV`.

(B). Given that,

`K.E'=65\ GeV`

We need to calculate the distance

Using formula of kinetic energy

`K.E=F* d`

Put the value into the formula

`65*10^(9)=2.1*10^(6)* d`

`d=(65*10^(9))/(2.1*10^(6))`

`d=30952.38\ m`

`d=30.95\ km`

The distance is 30.95 km.

(C). We know that,

The direction of force is taken the direction of electric field.

If the charge is negative then the direction of force is opposite direction of electric field.

So, The direction of electron is opposite direction of the electric field.

Hence, This is the required solution.