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. Answer:

    The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]

    Explanation:

    Given that,

    Electric field [tex]E=1.50\times10^{3}\ N/C[/tex]

    Distance = -0.0200

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

    Potential energy [tex]P.E=5.04\times10^{-17}\ J[/tex]

    Change in potential energy [tex]\Delta P.E=-9.60\times10^{-17}\ J[/tex] 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

    [tex]W=-eE\Delta x[/tex]

    Put the value into the formula

    [tex]W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))[/tex]

    [tex]W=-5.04\times10^{-17}\ J[/tex]

    We need to calculate the initial velocity

    Using change in kinetic energy,

    [tex]\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2[/tex]

    [tex]\Delta K.E=\dfrac{-3mv^2}{8}[/tex]

    Now, using work energy theorem

    [tex]\Delta K.E=W[/tex]

    [tex]\Delta K.E=\Delta U[/tex]

    So, [tex]\Delta U=W[/tex]

    Put the value in the equation

    [tex]\dfrac{-3mv^2}{8}=-5.04\times10^{-17}[/tex]

    [tex]v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}[/tex]

    Put the value of m

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

    [tex]v=1.21\times10^{7}\ m/s[/tex]

    We need to calculate the change in potential energy

    Using given potential energy

    [tex]\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})[/tex]

    [tex]\Delta U=-4.56\times10^{-17}\ J[/tex]

    We need to calculate the speed of electron

    Using change in energy

    [tex]\Delta U=-W=-\Delta K.E[/tex]

    [tex]\Delta K.E=\Delta U[/tex]

    [tex]\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}[/tex]

    Put the value into the formula

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

    [tex]v_{f}=1.5\times10^{7}\ m/s[/tex]

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

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