Answer to Question #250484 in Physics for Soriano

Question #250484
Particles A and B carrying charge 3.0nC are at the base corners of an equilateral triangle 2.8 m on a side

Calculate the electric potential energy for the triangular charge distribution?
1
Expert's answer
2021-10-13T09:07:56-0400

The electric potential energy of two point charges is given as follows:


W=kq1q2rW = k\dfrac{q_1q_2}{r}

where k=9×109Nm2/C2k = 9\times 10^{9}Nm^2/C^2 is the Coulomb constant, q1=q2q=3×109Cq_1 = q_2 \equiv q = 3\times 10^{-9}C are the charges (equal in this case), and r=2.8mr = 2.8m is the distance between the charges. Thus, obtain:


W=9×109(3×109)22.83.0×108JW = 9\times 10^{9}\cdot \dfrac{(3\times 10^{-9})^2}{2.8} \approx 3.0\times 10^{-8}J

ANswer. 3.0×108J3.0\times 10^{-8}J.


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