# 0 and negative electric potential in uniform plates

I had a question with two parallel conducting plates with voltages of +30V and -10V, with a separation of 0.2m. This got me thinking, at a certain point (0.15m from +30V plate) wouldn't the electric potential be 0? At the bottom (when a charge touches the -10V plate), wouldn't the electric potential be -10V?

I learnt electric potential as the amount of energy a charge would get if it was placed at that point, so it seems that at 0.15m the charge would not move yet intuitively I feel like it should because it's in an electric field. And at the bottom, the charge would have negative energy? Wouldn't it have 0 energy as the postive test charge unites with the negative plate?

The absolute value of the electric potential isn't used much, and in circuits is essentially arbitrary. All that matters is the difference in energy from one potential to another.

You might want to think about the analogous energy that mass has at various gravitational potentials. Say that we want to measure the energy we get from dropping a mass in our lab. We can measure the energy it has by looking at it's height above the lab table. We set the table to be $h=0$ and measure the mass to be $h=1m$. When it drops to the table, it releases energy equal to $mg \Delta h$

For other purposes, it might make more sense to measure the height above the floor of the lab. In that case, the block now has $h=2.42m$ and we drop it to the table at $h=1.42m.$ The energy released in both cases is identical.

Did the block have more energy when it was measured to be 2.42m above the lab floor than when it was measured to be 1m above the lab table? No. Those are just different references and we can arbitrarily set the zero point to whatever is convenient.

In the circuit, the same thing happens. We can set the 0V potential to be whatever is convenient. All that matters for understanding the circuit is the difference in potential between two points.

so it seems that at 0.15m the charge would not move yet intuitively I feel like it should because it's in an electric field

It is not the value of the potential you should be worried about it is the value of the potential gradient.

The fact that the potential is zero, or 17 or -54, is irrelevant, rather in terms of the force on the charge it is the rate of change of potential with position which gives you minus the electric field strength.

And at the bottom, the charge would have negative energy? Wouldn't it have 0 energy as the postive test charge unites with the negative plate?

You are mixing up two different situations.

If the positive test charge united with a negative charge on the bottom plate then the whole potential diagram would change as the potential difference between the two plates would change.

The test charge is not supposed to neutralise charges and in fact you should think of it as a charge which is infinitesimally small with the definition of the magnitude of an electric field as the limit of $\dfrac{\text{force on test charge}}{\text {magnitude of test charge}}$ as the magnitude of the test charge tends to zero.