Solution:

We are given three charges and need to calculate the potential energy of the system when the third charge $q = -4.20 \, \text{nC}$ is placed at point $b$.

Charges:

1. $q_1 = +9.80 \, \text{nC}$
2. $q_2 = -9.80 \, \text{nC}$
3. $q = -4.20 \, \text{nC}$ (placed at point $b$)

Geometry:

• The distance between $q_1$ and $q_2$ is $r_{12} = 8.00 \, \text{cm} = 0.080 \, \text{m}$.
• The distance between $q_1$ and $q$ (point $b$) is $r_{1b} = 4.00 \, \text{cm} = 0.040 \, \text{m}$.
• The distance between $q_2$ and $q$ (point $b$) is $r_{2b} = 4.00 \, \text{cm} = 0.040 \, \text{m}$.

Formula for Potential Energy:

The total potential energy of the system is given by:

$U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q}{r_{1b}} + \frac{q_2 q}{r_{2b}} \right)$

where:

• $k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$ (Coulomb's constant),
• $r_{12}$, $r_{1b}$, and $r_{2b}$ are the distances.

Step-by-step Calculation:

1. First Term: $\frac{q_1 q_2}{r_{12}}$ $\frac{q_1 q_2}{r_{12}} = \frac{(+9.80 \times 10^{-9})(-9.80 \times 10^{-9})}{0.080}$ $= -1.2005 \times 10^{-14} \, \text{J}$

2. Second Term: $\frac{q_1 q}{r_{1b}}$ $\frac{q_1 q}{r_{1b}} = \frac{(+9.80 \times 10^{-9})(-4.20 \times 10^{-9})}{0.040}$ $= -1.029 \times 10^{-14} \, \text{J}$

3. Third Term: $\frac{q_2 q}{r_{2b}}$ $\frac{q_2 q}{r_{2b}} = \frac{(-9.80 \times 10^{-9})(-4.20 \times 10^{-9})}{0.040}$ $= +1.029 \times 10^{-14} \, \text{J}$

4. Total Potential Energy: Adding all terms: $U = k \left( -1.2005 \times 10^{-14} - 1.029 \times 10^{-14} + 1.029 \times 10^{-14} \right)$ $U = k \times (-1.2005 \times 10^{-14})$ Substituting $k = 8.99 \times 10^9$: $U = (8.99 \times 10^9)(-1.2005 \times 10^{-14})$ $U = -1.08 \, \mu \text{J}$

Final Answer:

The total potential energy of the system is: $U = -1.08 \, \mu \text{J}$

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Follow-up Questions:

1. How is potential energy influenced by the sign of charges in the system?
2. What happens to the potential energy if all charges are positive?
3. How would the result change if the distances were doubled?
4. How is the total potential energy affected if $q = +4.20 \, \text{nC}$ instead of $-4.20 \, \text{nC}$?
5. How does the principle of superposition apply to calculating potential energy?

Tip:

Remember, the potential energy is scalar, so always consider the distances and the product of charges carefully to include their signs.