Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J
The potential energy of this two particle system relative to the potential energy at infinite separation is [tex]\bold {-32.5x 10^-^5\ J}[/tex].
The potential energy of this system of charges,
[tex]\bold {Ue = k\dfrac{q1q2}{r}}[/tex]
Where;
k - Coulumb's constant
q1 and q2 - magnitudes of the charges
r - distance between the charges
Put the values in the equation,
[tex]\bold {Ue = 9.0x10^9\times \dfrac {5.5 x 10^{-8} C \times -2.3 x10^{-8} C}{3.5 \times 10^{-2}}}\\\\\bold {Ue= -32.5 x 10^-^5\ J}[/tex]
Therefore, the the potential energy of this two particle system relative to the potential energy at infinite separation is [tex]\bold {-32.5x 10^-^5\ J}[/tex].
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