# Difference between voltage, electrical potential and potential difference

I am having hard time to visualize these two concepts in my mind seriously.

First of this confusion came from two parallel plates that was connected to a power supply, charged then disconnected from power supply and then separated from each other, strangely potential difference increased but why? I have learn that the electrical potential is$$V= k \frac{q}{d}$$and when the distance increases the potential of a point must drop but why when we are talking for potential difference it is increasing, doesn't every point between these plates feel less stress as the plates moves apart and doesn't this mean potential is dropping and so the potential difference as well?

There is in fact a mistake in your argument. This is an error which I believe is made by inquisitive students and one which helps the student learn electrostatics better. I'm glad that you've asked this. V = kq/d isn't applicable everywhere. It's only applicable to a static point charge or a outside a uniform sphere. But it's not true for all situations. An example would be a system of parallel plates.

What is instead true is that V is the integral of electric field with respect to small distance. This is always true, as it's the definition of V. In simpler words, V is the work required to bring a unit positive charge from infinity to the position it's in presently.

Now if you have a system of capacitors, Q = CV. This comes from the very definition of capacitance. Now for a parallel plate capacitor, C is proportional to the area of the plates and inversely proportional to distance between the plates. When you pull two plates apart, the area of the plates stays constant. Since there's nowhere the charge can go , the charge on each plate stays constant too. The only thing that changes is the distance. As V = Q/C, this would mean that as 1/C is proportional to distance, V is proportional to the distance between the plates too. So when you increase the distance, you in fact are increasing the potential difference. I hope I managed to explain it. Have fun learning !

I'll give a simple explanation. Potential difference is the difference of potential between 2 points (say A and B). Potential of any point would be the potential difference between that point and another point C, the specialty of C is that its potential is assumed to be 0.Usually we keep C at infinity.

In your question, we have a capacitor, we give it some charge , we disconnect it , now the charge remains same,meaning the electric field between them is same as before.You know that $$V=E*d$$ where V is the potential difference , E is the electric field and d is the distance. It is given that d increases,hence V increases.

$$V=\frac{Kq}{d}$$ is the potential of a point at a distance d from a point charge, in fact even here what we are doing is simply V=Ed but since the electric field changes wrt time be integrate($$\int E.dx$$) and get the relation $$V=\frac{Kq}{d}$$. As a conclusion, $$V=\frac{Kq}{d}$$ is applicable only for point charges/spheres(if d>radius) and V=$$\int E.dx$$ holds everywhere.

You are mixing the properties of a point charge and that of a capacitor.

The formula you've written works perfectly for point charges and intuitively for two entities holding the same nature of charge and to be very precise, you should use $$r$$ instead of $$d$$ when dealing with point charges as the electric field from a point charge has spherical symmetry. Note that this formula won't work for charged plates. It makes sense that when two positive point charges of same polarity (not necessarily of same magnitude) are moved farther from each other the potential energy of the system decreases. But in your question you are talking about what are known as capacitors in the business. You might be knowing that when two oppositely charged plates are moved away from each other the potential energy of the system increases as you are doing work against the electric field and a capacitor comprises two plates placed in close proximity which hold equal and opposite charges. So here, the potential difference between the plates is given by $$E=\frac{V}{d}$$ As you might be knowing that the electric field in the region between the two plates is always constant no matter how far you move the plates provided the plates are parallel to each other and one must perfectly block the view of the other. This way the ratio $$\frac{V}{d}$$ will always be constant. So you may conclude that whenever distance between the plates increases the potential energy of the system also increases.

• So as I understand the reason is that the electrical field is constant due to parallel plates and I think there is a proof for that, that includes some integrals in it so okey I can accept that with a promise I ll learn what's behind this aswell but something other makes me think and that is, if we increase the distance between charged plates while they are still connected to battery because the potential difference is constant due to battery will electrical field(E) decrease? If so why is that? Mar 27, 2019 at 13:24
• @E.Berk The field lines will decrease as the battery will have a hard time pushing electrons into the negative plate as the electrons won't want to leave the positive plate and the already negative plate will be hating more electrons. Mar 27, 2019 at 13:28

In a parallel plate capacitor, the electric field is homogeneous between the plates (we neglect boundary effects here) and given by $$E = \frac{\sigma}{\epsilon\epsilon_0},$$ where $$\epsilon$$ is the relative dielectric constant of the medium between the plates, and $$\sigma$$ is the charge density on one plate. The direction of the electric field vector is normal to the surface of the plates. The formula follows from a direct application of Gauss' law of electrostatics.

The potential difference between the plates is the line integral of $$E$$ along a line connecting the two plates, giving $$\Delta V = \frac{\sigma}{\epsilon\epsilon_0}\times d.\quad (1)$$ Here, $$d$$ is the separation of the two plates.

If you charge the plates by connecting them to a voltage source supplying the voltage $$\Delta V$$, then you get the surface charge density of magnitude $$\sigma = \frac{\epsilon\epsilon_0}{d}\Delta V$$ on each plate (the sign of the charge density will be different on the two plates).

Disconnecting the plates from the voltage source leaves the charges on the plates unchanged. Likewise, the charge density $$\sigma$$ does not change, if you further increase the separation $$d$$ of the plates. The voltage will increase with $$d$$ according to eq. (1).

In fact, the quantity $$\epsilon\epsilon_0/d$$ is called the capacitance per unit area of the parallel plate system.

For parallel plates where the separation $$d$$ is much smaller than the dimensions of the plates (diameter for circular plates), and the potential difference between the plates is $$V$$, the electric field $$E$$ is given by

$$E=\frac{V}{d}$$

Where the electric field $$E$$ is directed from the + plate to the – plate.

Capacitance, $$C$$ is electrically defined as the amount of charge $$q$$ on the plates per volt across the plates, or

$$C=\frac{q}{V}$$

In terms of the physical characteristics of a capacitor, the capacitance is given by $$C=\frac{εA}{d}$$

Equating the last two equations gives us

$$V=\frac{qd}{εA}$$

Where $$ε$$ is the electrical per permittivity of the medium between the plates.

Substituting $$V$$ from the last equation for $$V$$ in the first equation gives us

$$E=\frac{q}{εA}$$

The last equation shows that the electric field strength between the plates does not depend on the plate separation. Now returning to the first equation expressed in terms of potential difference we have

$$V=Ed$$

Since $$E$$ is constant, increasing the separation increases the electrical potential difference. This makes sense when you consider the following definition of potential difference, or voltage:

The potential difference, $$V$$, is defined as the work (joules) per unit charge $$q$$ (coulomb) required to move the charge between the points.

The force on a charge $$q$$ between the plates is $$qE$$. The work required to move the charge from one plate to another is

$$W=qEd$$ The work per unit charge is $$\frac{W}{q}=Ed=V$$

Hope this helps.

• Can you please explain where did the ''d'' go in the capacitance equations when we equated them? Because I didn't see something we can divide it. (Edit) I did the same math I think you divide it early, after equating it to E=V/d I see they cancel each other :D Mar 27, 2019 at 15:00
• @E.Berk Oops sorry. I obviously inadvertently omitted it. See correction. Mar 27, 2019 at 15:06
• @E.Berk My first goal was to show that the field is constant and independent of the separation. That justifies holding it constant when increasing the plate separation and showing that the potential increases with separation. My second goal, since the increase in potential seems sort of counterintuitive, was to show that increase in potential means more work to move the charge a further distance against a constant force. In any case, I hope it was of some help. Mar 27, 2019 at 15:16