Let's say I have two points in a non-uniform electric field and those points have the electric potential 10 V and 5 V. If I then would use a voltmeter to determine the voltage between those points would the voltage be turn out to be 10-5= 5 V?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Are you assuming an ideal voltmeter ('infinite' input impedance)? Also, are you assuming that the red (positive) lead goes to the point with the higher potential? $\endgroup$– Alfred CentauriCommented Nov 12, 2018 at 1:26
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
Yes. The uniformity, or lack thereof, of the electric field does not impact the potential difference between two points in the slightest.
answered Nov 12, 2018 at 0:05
-
$\begingroup$ Just to add to your correct answer, the electric field is only how fast the electric potential is changing in space. $E_x=dV/dx$ It doesn’t change that the total electric potential difference between those points is 5V. $\endgroup$– BandooCommented Mar 12, 2020 at 4:50
$\begingroup$
$\endgroup$
It depends on whether the potential difference of the two points stays the same after you connect the voltmeter. Usually in practice of electrical circuits that is the case, because the voltage source is capable to sustain the potential difference.
However, there are situations where the voltmeter will cause the potential difference to drop and then it will show value different from the actual potential difference before the voltmeter was connected. This happens when the source of the potential difference is so soft that a small current flowing through the voltmeter will upset the distribution of electric charge.
For example, when AC transformer has primary coil connected to AC grid but its secondary coil's poles are not connected to any circuit, potential of one pole of secondary will oscillate somewhere around the potential of the ground, while potential of the ground is constant. But as soon as you connect voltmeter to both ground and the pole to measure their potential difference, this state will change and the connected pole will acquire the same potential as the ground. The voltmeter will show correctly 0 volts.
answered Nov 12, 2018 at 2:32