# High potential energy but low potential

I'm really struggling to understand electric potential. My textbook makes this example:

Could you please elaborate a little more one why the potential at point $$a$$ is higher than the potential at point $$b$$? Is it because the negative plate as been taken as zero potential? If the zero potential is chosen arbitrarily, what would happen if I'd consider the positive plate as the zero? Could you help me clarify this please?

Could you please elaborate a little more one why the potential at point $$a$$ is higher than the potential at point $$b$$?

The definition of high and low potential is based on positive electric charge.

The potential at $$a$$ is higher than at $$b$$ because it takes work by an external agent to move positive charge from the negative plate to the positive plate against the direction of the repulsive force exerted by the electric field $$E$$.

The force exerted by the electric field (assumed constant between the plates) is

$$F=QE$$

If the distance between the plates is $$d$$, then the work done moving the charge from the negative to positive plate giving the charge potential energy is

$$W=Fd=QEd$$

The potential difference, $$V_{ab}$$ between the plates is related to the electric field by

$$E=\frac{V_{ab}}{d}$$

Therefore, the work and potential energy can be written as

$$W=QV_{ab}$$

The gravitational analogy is lifting an object a height $$h$$ against the downward force of the gravitational field. The gravitational field near the surface of the earth is $$g$$ and the gravitational force is $$F=mg$$. When lifted the height $$h$$ its gravitational potential energy is $$mgh$$.

Is it because the negative plate as been taken as zero potential?

No, it is because a negatively charged plate will always be at lower potential than a positively charged plate. The negative plate does not have to be taken as zero potential. But whatever potential it is it has to be lower than the potential of the positive plate because it takes work to move positive charge against the direction of the electric field, which is defined as the direction of the force that a positive charge would experience if placed in the field. The work moving the charge from the negative to positive plate increases both the potential and potential energy of the positive charge.

what would happen if I'd consider the positive plate as the zero?

If you did assign it zero potential, then the potential of the negative plate would have to be some negative value so that $$V_{a}>V_{b}$$ and $$V_{ab}=V_{a}-V_{b}>0$$.

For example, if $$V_{a}=0$$ volts, then $$V_{b}<$$ 0, for example -5 volts. Then $$V_{a}-V_{b}=0-(-5)=+5$$ volts

I think I understand, the definition of $$V_{ab}$$ is $$V_{ab}=V_{a}-V_{b}=(U_{a}-U_{b})q$$, since in this case, as the book says too, $$U_{a}-U_{b}<0$$ and $$q<0$$, $$V_{a}-V_{b}>0$$, and therefore $$V_{a}>V_b$$. Correct right?

You got it. The key is the potential energy of the negative charge ($$q<0$$) is lower at $$a$$ than at $$b$$ whereas the potential energy of the positive charge ($$q>0$$) is higher at $$a$$ than at $$b$$. I think this should also help you regarding your prior, similar question.

Can I ask you one more thing please? The books says that $$V_{a}-V_{b}= -W_{ab}/q$$, where $$W_{ab}$$ is the work done by the electric field as the charge moves from 𝑏 to 𝑎

Yes. As I said in the beginning, because the potential is greater at $$a$$ than $$b$$ an external (to the field) agent is needed to do work to move positive charge from $$b$$ to $$a$$. Since the direction of the external force is the same as the direction of movement (displacement) of the charge, the work it does is positive. The gravity analogy is an external force (e.g., you) lifting an object to increase its gravitational potential and gravitational potential energy.

At the same time the external force is doing positive work, the electric field is doing an equal amount of negative work when the positive charge moves from $$b$$ to $$a$$. This is because the force of the electric field is opposite the direction of movement (displacement) of the positive charge. In doing negative work the electric field takes the energy supplied by the external agent and stores it as electrical potential energy of the charge/field system.

and then it says that the potentials $$V_a$$ and $$V_b$$ (or the electric potentials in general) are due to the charges on the plates, not due to the test charge between the plates.

That's correct, as long as the test charges remain between the plates and are not allowed to combine with charges on the plates. If the figure, if the negative charge is released and allowed to combine with the positive charges on the positive plate the net positive charge on the positive plate will be reduced, altering the electric field and electric potential between the plates.

Is this because:$$V_{a}-V_{b}=-W_{ab}/q=-Eqd/q=-Ed$$ and therefore since the electric field depends only on the charge that produces the field, $$V_{ab}$$ depends only on the charges on the plates? –

Yes, because the $$q$$ in the equation is the test charge $$q$$, and not the $$q$$ on the plates.

Hope this helps.

• Thanks for the answer! I think i understand, the definition of $V_{ab}$ is $V_{ab}=V_a-V_b=(U_a-U_b)/q$, since in this case, as the book says too, $U_a-U_b<0$, and $q<0$, $V_a-V_b>0$, and therefore $V_a>V_b$. Correct right? Mar 17, 2021 at 14:12
• @AndreaBurgio Yep. See update to my answer. Mar 17, 2021 at 14:34
• Thank you so much! Can i ask you one more thing please? The books says that $V_a-V_b=-W_{ab}/q$, where $W_{ab}$ is the work done by the electric field as the charge moves from $b$ to $a$ and then it says that the potentials $V_a$ and $V_b$ (or the electric potentials in general) are due to the charges on the plates, not due to the test charge between the plates. Is this because: $V_a-V_b=-W_{ab}/q=-Eqd/q=-Ed$ and therefore since the electric field depends only on the charge that produces the field, $V_{ab}$ depends only on the charges on the plates? Mar 17, 2021 at 15:09
$$\vec{E}=-\vec{\nabla}V.$$