Question

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint.

Answer 1

The work done by the Coulomb force will be "$6.08\times 10^{-21} \ J$".

Let us define the required work done to move that alpha particle to the one of the mid point of the side length as follows.

→ $W = \frac{4kqQ}{r_1} -(\frac{2kqQ}{r_2} +\frac{2kqQ}{r_3} )$

→      $=2kqQ(\frac{2}{r_1} -\frac{1}{r_2} -\frac{1}{r_3} )$

→      $=2(8.99\times 10^9)(-1.6\times 10^{-19})(2)(1.6\times 10^{-19})$

→      $= (\frac{2}{\sqrt{(5\times 10^{-9})^2+(5\times 10^{-9})} } -\frac{1}{(5\times 10^{-9})} -\frac{1}{\sqrt{(10\times 10^{-9})^2} +(5\times 10^{-9})^2} )$

→      $= 6.08\times 10^{-21} \ J$

Thus the above answer is appropriate.  

Learn more about work done here:

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