What is the escape speed of an electron launched from the surface of a 1.5 cm diameter glass...

Question:

What is the escape speed of an electron launched from the surface of a 1.5 cm diameter glass sphere that has been charged to 10 nC?

Potential energy:

The term potential energy can define as the accumulated energy in the object due to the position corresponding to the other objects. The potential energy of a body in a vertical motion directly alters with height.

Answer and Explanation:

Given data:

  • The sphere glass diameter is {eq}d = 1.5\,{\rm{cm}} = 1.5 \times {10^{ - 2}}\,{\rm{m}} {/eq}
  • The charge is {eq}q = 10\,{\rm{nC}} = 10 \times {10^{ - 9}}\,{\rm{C}} {/eq}


The expression for the radius of the sphere glass is

{eq}r = \dfrac{d}{2} {/eq}


Substituting the values in the above equation as,

{eq}\begin{align*} r &= \dfrac{d}{2}\\ r &= \dfrac{{1.5 \times {{10}^{ - 2}}}}{2} \end{align*} {/eq}

The expression for the potential of the sphere is

{eq}V = \dfrac{{kq}}{r} {/eq}


Substituting the values in the above equation as,

{eq}\begin{align*} V &= \dfrac{{kq}}{r}\\ V &= \dfrac{{\left( {9 \times {{10}^9}} \right)\left( {10 \times {{10}^{ - 9}}} \right)}}{{\left( {\dfrac{{1.5 \times {{10}^{ - 2}}}}{2}} \right)}}\\ V &= 12000\,{\rm{J}} \end{align*} {/eq}


The expression for the potential energy of the electron is

{eq}u = {q_e}V {/eq}

  • Here {eq}{q_e} = 1.6 \times {10^{ - 19}}\,{\rm{C}} {/eq} is the charge of the electron.


Substituting the values in the above equation as,

{eq}\begin{align*} u &= {q_e}V\\ u &= \left( {1.6 \times {{10}^{ - 19}}} \right)\left( {12000} \right)\\ u &= 1.92 \times {10^{ - 15}}\,{\rm{J}} \end{align*} {/eq}


The expression for the conservation of energy is

{eq}k.e = u {/eq}

  • Here {eq}k.e = \dfrac{1}{2}m{v^2} {/eq} is the kinetic energy.
  • Here {eq}m = 9.1 \times {10^{ - 31}}\,{\rm{kg}} {/eq} is the mass of the electron.


Substituting the values in the above equation as,

{eq}\begin{align*} k.e &= u\\ \dfrac{1}{2}m{v^2} &= 1.92 \times {10^{ - 15}}\\ v &= \sqrt {\dfrac{{1.92 \times {{10}^{ - 15}} \times 2}}{{9.1 \times {{10}^{ - 31}}}}} \\ v &= 6.49 \times {10^7}\,{\rm{m/s}} \end{align*} {/eq}

Thus the escape velocity is {eq}v = 6.49 \times {10^7}\,{\rm{m/s}} {/eq}


Learn more about this topic:

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