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- Thread starter Tony11235
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Physics Monkey

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[tex]

\phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}

[/tex]

assuming the test charges are truly infinitesimal (i.e. we neglect their contribution).

The electrostatic potential energy (or just energy) is

[tex]

E_{q}= \frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}

[/tex]

in the case of a positive test charge and

[tex]

E_{-q}= -\frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}

[/tex]

in the case of a negative test charge. The energy or electrostatic potential energy is different in each case, but the electrostatic potential of the charge Q is the same in both cases. It's mostly a matter of terminology.

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Physics Monkey said:

[tex]

\phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}

[/tex]

assuming the test charges are truly infinitesimal (i.e. we neglect their contribution).

The electrostatic potential energy (or just energy) is

[tex]

E_{q}= \frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}

[/tex]

in the case of a positive test charge and

[tex]

E_{-q}= -\frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}

[/tex]

in the case of a negative test charge. The energy or electrostatic potential energy is different in each case, but the electrostatic potential of the charge Q is the same in both cases. It's mostly a matter of terminology.

So to clear this up, the electrostatic potential energy of the positive test charge is greater?

- #4

Physics Monkey

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Yes. It has positive energy while the negative charge has negative energy.

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