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

Answer:

The speed of electron is $1.5\times10^{7}\ m/s$

Explanation:

Given that,

Electric field $E=1.50\times10^{3}\ N/C$

Distance = -0.0200

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

Potential energy $P.E=5.04\times10^{-17}\ J$

Change in potential energy $\Delta P.E=-9.60\times10^{-17}\ J$ 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

$W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))$

$W=-5.04\times10^{-17}\ J$

We need to calculate the initial velocity

Using change in kinetic energy,

$\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2$

$\Delta K.E=\dfrac{-3mv^2}{8}$

Now, using work energy theorem

$\Delta K.E=W$

$\Delta K.E=\Delta U$

So, $\Delta U=W$

Put the value in the equation

$\dfrac{-3mv^2}{8}=-5.04\times10^{-17}$

$v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}$

Put the value of m

$v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}$

$v=1.21\times10^{7}\ m/s$

We need to calculate the change in potential energy

Using given potential energy

$\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})$

$\Delta U=-4.56\times10^{-17}\ J$

We need to calculate the speed of electron

Using change in energy

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

$\Delta K.E=\Delta U$

$\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}$

Put the value into the formula

$v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}$

$v_{f}=1.5\times10^{7}\ m/s$

Hence, The speed of electron is $1.5\times10^{7}\ m/s$