Question

An electron has an initial speed of 5.85 106 m/s in a uniform 5.55 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? opposite direction to the electron's initial velocity same direction as the electron's initial velocity not enough information to decide (b) How far does the electron travel before coming to rest? m (c) How long does it take the electron to come to rest? s (d) What is the electron's speed when it returns to its starting point? m/s

Answer 1

(a) same direction as the electron's initial velocity

The direction of the acceleration is opposite to the direction of the velocity of the electron. This means that the electron is feeling a repulsive force, in a direction opposite to its initial velocity.

For a negative charge, we know that the electrostatic force and the electric field have opposite directions, because in the formula

`F=qE`

q is negative. Therefore, the electric field must be in the same direction as the initial velocity of the electron.

(b)
`1.76\cdot 10^(-4)m`

When the electron comes to rest, all its initial kinetic energy has been converted into electric potential energy. So we can write

`K = \Delta U`

`(1)/(2)mv^2= qEd`

where

`m=9.11\cdot 10^(-31) kg` is the electron mass

`v=5.85\cdot 10^6 m/s` is the electron initial speed

`q=1.6\cdot 10^(-19)C` is the magnitude of the electron charge

`E=5.55\cdot 10^5 N/C` is the electric field

`d` is the distance covered

Solving the equation for d, we find

`d=(mv^2)/(2qE)=((9.11\cdot 10^(-31) kg)(5.85\cdot 10^6 m/s)^2)/(2(1.6\cdot 10^(-19)C)(5.55\cdot 10^5 N/C))=1.76\cdot 10^(-4)m`

which corresponds to 0.17 mm.

(c)
`6\cdot 10^(-11) s`

First of all, we need to find the electrostatic force acting on the electron:

`F=qE=(-1.6\cdot 10^(-16)C)(5.55\cdot 10^5 N/C)=-8.88\cdot 10^(-14) N`

Now we can find the acceleration of the electron:

`a=(F)/(m)=(-8.88\cdot 10^(14) N)/(9.11\cdot 10^(-31) kg)=-9.75\cdot 10^(16) m/s^2`

(the acceleration is negative because it is opposite to the electron's direction of motion)

And now we can find the time taken for the electron to stop to a velocity of v=0 starting from
`u=5.85\cdot 10^6 m/s`:

`a=(v-u)/(t)\\t=(v-u)/(a)=(0-(5.85\cdot 10^6 m/s))/(-9.75\cdot 10^(16) m/s^2)=6\cdot 10^(-11) s`

(d)
`5.85\cdot 10^6 m/s`

When it returns to the starting point, all the electric potential energy gained by the electron through the distance d will be re-converted back into kinetic energy. If there is no loss of energy, therefore, this means that the electron will have the same kinetic energy it had at the beginning of the motion: therefore, its speed will be equal to its initial speed,
`5.85\cdot 10^6 m/s`.