Question

A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −0.0200 m in the same direction as the electric field. The electron's speed has fallen by half when it reaches x = 0.190 m, a change in potential energy of 5.04 ✕ 10−17 J. The electron continues to x = −0.210 m within the constant electric field. If there's a change in potential energy of −9.60 ✕ 10−17 J as it goes from x = 0.190 m to x = −0.210 m, find the electron's speed (in m/s) at x = −0.210 m.

Answer 1

The speed of electron is
`1.5*10^(7)\ m/s`

Step-by-step explanation:

Given that,

Electric field
`E=1.50*10^(3)\ N/C`

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy
`P.E=5.04*10^(-17)\ J`

Change in potential energy
`\Delta P.E=-9.60*10^(-17)\ J` as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

`W=-eE\Delta x`

Put the value into the formula

`W=-1.6*10^(-19)*1.50*10^(3)*(0.190-(-0.0200))`

`W=-5.04*10^(-17)\ J`

We need to calculate the initial velocity

Using change in kinetic energy,

`\Delta K.E = (1)/(2)m((v)/(2))^2+(1)/(2)mv^2`

`\Delta K.E=(-3mv^2)/(8)`

Now, using work energy theorem

`\Delta K.E=W`

`\Delta K.E=\Delta U`

So,
`\Delta U=W`

Put the value in the equation

`(-3mv^2)/(8)=-5.04*10^(-17)`

`v^2=(8*(5.04*10^(-17)))/(3m)`

Put the value of m

`v=\sqrt{(8*(5.04*10^(-17)))/(3*9.1*10^(-31))}`

`v=1.21*10^(7)\ m/s`

We need to calculate the change in potential energy

Using given potential energy

`\Delta U=-9.60*10^(-17)-(-5.04*10^(-17))`

`\Delta U=-4.56*10^(-17)\ J`

We need to calculate the speed of electron

Using change in energy

`\Delta U=-W=-\Delta K.E`

`\Delta K.E=\Delta U`

`(1)/(2)m(v_(f)^2-v_(i)^2)=4.56*10^(-17)`

Put the value into the formula

`v_(f)=\sqrt{(2*4.56*10^(-17))/(9.1*10^(-31))+(1.21*10^(7))^2}`

`v_(f)=1.5*10^(7)\ m/s`

Hence, The speed of electron is
`1.5*10^(7)\ m/s`