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I know that when you have a charge, it creates an electric field that is equipotential. So, can you have potential difference without adding another charge? Maybe putting a material across the electric field.

Formula for electric potential at one point:

$$V(Q,r) = \frac{1}{4\pi\epsilon_0} \times \frac{Q}{r^2}$$

Being \$\epsilon_0\$ the permittivity constant for the vacuum

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    \$\begingroup\$ The universe, or an endless empty, has a potential of 0. So even if you had an infinite empty space around the single charge, you'd still have an electrical field diminishing toward 0 with increasing r. Of course, you cannot observe the field without introducing another charge to disturb the symmetry and cause a net force. \$\endgroup\$
    – JimmyB
    Commented Nov 11, 2016 at 15:18

2 Answers 2

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Yes

Since Q=CV, one can change the dielectric material ( and thus C )and the voltage and field E will change . Assuming lossless dielectric , charge is constant before and after.

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Q=CV, so if the point charge has capacitance then there will be a voltage. But, I hear you possibly murmuring, "what capacitance can a point source have". Well, an infinitely small point doesn't have capacitance but any finite object does. Consider the capacitance of two concentric spheres and then let the outer sphere become infinitely large: -

enter image description here

This picture tells you what the capacitance is between concentric spheres and if the outer sphere went to infinity, there is still capacitance so yes, for a non-infinitely small charged object there will be a voltage.

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