Chapter 4: Q8E (page 777)

Three equal 1.20 µC point charges are placed at the corners of an equilateral triangle with sides 0.400 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Short Answer

Expert verified

The potential energy of the system is 0.097 J_.

Step by step solution

Potential energy formula

The potential energy of a body is given by

$U = \frac{1}{4 {ττε}_{0}} \sum_{i < j} \frac{q_{i} q_{j}}{r_{ij}}$

Consider the three charges arethe same so the potential energy will be as follows:

$U = \frac{1}{4 {πε}_{0}} \sum_{I < J} \frac{q_{i} q_{j}}{r_{i j}} U = \frac{1}{4 {πε}_{0}} (\frac{q_{i} q_{j}}{r_{12}} + \frac{q_{2} q_{3}}{r_{23}} + \frac{q_{1} q_{3}}{r_{13}}) U = \frac{1}{4 {ττε}_{0}} \frac{3 q^{2}}{r}$

Here, $ε_{0}$ is the permittivity of the free space.

Here, $\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} \frac{N . m^{2}}{C^{2}}$ .

Determine the potential energy of the system

Substitute the values and solve as:

$U = (9 \times 10^{9} \frac{N . m^{2}}{C^{2}}) (\frac{3 {(1.20 \times 10^{- 6} C)}^{2}}{0.4 m}) = 0.097 J$

Therefore, the potential energy of the system is 0.097 J

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Three equal 1.20 µC point charges are placed at the corners of an equilateral triangle with sides 0.400 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Short Answer

Step by step solution

Potential energy formula

Determine the potential energy of the system

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