Answer:
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That will be
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<em>=</em><em> </em><em><u>1</u></em><em><u>0</u></em><em><u>1</u></em><em><u>,</u></em><em><u>2</u></em><em><u>5</u></em><em><u>0</u></em><em><u>,</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>
Answer:
1.38 x 10^-18 J
Explanation:
q = - 1.6 x 10^-19 C
d = 5 x 10^-10 m
the potential energy of the system gives the value of work done
The formula for the potential energy is given by
![U =\frac{Kq_{1}q_{2}}{d}](https://tex.z-dn.net/?f=U%20%3D%5Cfrac%7BKq_%7B1%7Dq_%7B2%7D%7D%7Bd%7D)
So, the total potential energy of teh system is
![U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}](https://tex.z-dn.net/?f=U%20%3D%5Cfrac%7BKq_%7B1%7Dq_%7B2%7D%7D%7Bd%7D%2B%5Cfrac%7BKq_%7B2%7Dq_%7B3%7D%7D%7Bd%7D%2B%5Cfrac%7BKq_%7B1%7Dq_%7B3%7D%7D%7Bd%7D)
As all the charges are same and the distance between the two charges is same so the total potential energy becomes
![U =3\times \frac{Kq^{2}}{d}](https://tex.z-dn.net/?f=U%20%3D3%5Ctimes%20%5Cfrac%7BKq%5E%7B2%7D%7D%7Bd%7D)
K = 9 x 10^9 Nm^2/C^2
By substituting the values
![U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}](https://tex.z-dn.net/?f=U%20%3D3%5Ctimes%20%5Cfrac%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes%20%5C%201.6%20%5Ctimes%201.6%20%5Ctimes%2010%5E%7B-38%7D%7D%7B5%5Ctimes%2010%5E%7B-10%7D%7D)
U = 1.38 x 10^-18 J