# 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 i

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.

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1. Tryphena

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 $$1.5\times10^{7}\ m/s$$