High School

Two point charges (+9.7 nC and -9.7 nC) are located 8.00 cm apart. Let [tex]U = 0[/tex] when all of the charges are separated by infinite distances.

What is the potential energy if a third point charge [tex]q = -4.20 \text{ nC}[/tex] is placed at point C?

Given:
- Distance between charges: 8.00 cm
- Charge values:
- [tex]+9.7 \text{ nC}[/tex]
- [tex]-9.7 \text{ nC}[/tex]
- [tex]-4.20 \text{ nC}[/tex]

Note: The additional context provided in the question (e.g., "8.00 cm 8.00 cm b a +4.00 +4.00 +4.00") seems to be unrelated or erroneous. Please clarify if specific positions or configurations need to be considered.

Answer :

The potential energy of the system when a third point charge q = -4.20 nC is placed at point C is -5.53 x 10^-6 J.

The potential energy of a system of point charges is given by the equation U = k(q1q2)/r, where U is the potential energy, k is the Coulomb constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the separation distance between the charges. In this case, we have two charges (+9.7 nC and -9.7 nC) located 8.00 cm apart.

The potential energy between these two charges is zero since the charges are equal in magnitude and opposite in sign. When a third charge q = -4.20 nC is placed at point C, the potential energy is calculated by substituting the values into the equation. The result is -5.53 x 10^-6 J. The negative sign indicates that the system is in a stable configuration.


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