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I have a problem in where the electric potential has a constant value of six volts everywhere in a 3D region, the points are spread out. My question is would the electric potential energy of the system be the same all throughout no matter the distance since the electric potential is the same at all points?

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  • $\begingroup$ What system as there can be no charges in a region where the potential is constant? $\endgroup$ – Farcher Feb 13 '17 at 8:06
  • $\begingroup$ $$\mathbf{E}=-\nabla \phi$$ As long as the test charge is within the region, the charge experiences no electrical force and hence no transfer of electrical energy takes place. $\endgroup$ – Ng Chung Tak Feb 13 '17 at 12:47
  • $\begingroup$ There is a thing which you need to take an account ,what you refer as a reference point ,for calculation potential energy of V=6 change in potential energy is same $\endgroup$ – yuvraj singh Aug 23 at 10:45
  • $\begingroup$ That is zero if v=6 is constant with magnitude of test charge Q $\endgroup$ – yuvraj singh Aug 23 at 10:47
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Since potential is just work done or energy per unit charge, potential energy will also be same everywhere for a charge Q.

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  • $\begingroup$ so the electric potential energy will be the same everywhere right but would it be zero? or not ? $\endgroup$ – Dee Feb 13 '17 at 14:10
  • $\begingroup$ @Dee It can be zero, when charge is zero ie, field becomes zero. But constant potential energy means the change in potential energy will be zero, since its constant everywhere. $\endgroup$ – Allen Feb 14 '17 at 16:08
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What you want is

$$\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t}$$

The energy you might be looking for is: \begin{align} E &= \epsilon_0 \int\ \mathbf{E} \cdot \mathbf{E}\ \mathrm dV \\ &= \epsilon_0 \int\ \left(\nabla \phi + \frac{\partial \mathbf{A}}{\partial t}\right)\left(\nabla \phi + \frac{\partial \mathbf{A}}{\partial t} \right)\ \mathrm dV \end{align}

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