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
[tex]U =\frac{Kq_{1}q_{2}}{d}[/tex]
So, the total potential energy of teh system is
[tex]U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}[/tex]
As all the charges are same and the distance between the two charges is same so the total potential energy becomes
[tex]U =3\times \frac{Kq^{2}}{d}[/tex]
K = 9 x 10^9 Nm^2/C^2
By substituting the values
[tex]U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}[/tex]
U = 1.38 x 10^-18 J
The work which must be done to bring three electrons from a great distance apart is 5.0×10⁻¹⁰m from one another is 1.38 x 10⁻¹⁸ J
q = -1.6 x 10⁻¹⁹C
d = 5 x 10⁻¹⁰m
Potential energy of the system = Value of work done
Potential energy= U = Kq₁q₂/d
It is an equilateral triangle so the the charges and the distance between
the charges are the same.
Potential energy = 3 ₓ kq² / d
Substitute the values into the equation
3 × (9 × 10⁹ ˣ 1.6 ˣ 1.6 ˣ 10⁻³⁸) / (5ˣ 10⁻¹⁰)
U = 1.38 x 10⁻¹⁸ J
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