Change electric potential along uniform electric field

Question

I have a doubt. I do know that the potential at a point decreases as we move in the direction of electric field because $$V=\frac{KQ}{r}.$$

This was taught to me with a point change as an example. I understood that. But I don't understand how the potential reduces the same way in a uniform electric field. I find it difficult because electric field magnitude is same everywhere in a uniform electric field. And considering the following equation, I think the potential has to increase $$E=\frac{V}{R}$$ $$V=ER$$ But I later found that it is not the case. So here's how I thought about it

From the picture, it can be seen that a uniform electric field is generated by a sheet of uniform charge density. So now I thought that $Q$ (charge) is generating the first field Line , and another charge is generating the second field Line. That way A happens to be closer to $Q$ than B. So $V_\mathrm A$ is greater than $V_\mathrm B$.

Is it the right way? Can you tell me?

Answer 1

And considering the following equation, I think the potential has to increase: $$E = \frac{V}{R}$$ $$V = ER$$

This equation is incorrect. The correct equation is: $$ \textbf{E}= - \textbf{grad}(V)$$ or equivalently, for a small displacement $d\textbf{s}$:

$$∆V = -\int_{i}^{f}{\textbf{E}.d\textbf{s}}$$

Both these equations mathematically suggest that the potential reduces as you go along the direction of the electric field. In the case that the field is uniform, you can take $\textbf{E}$ out of the integral, to get: $$∆V = -\textbf{E}.∆\textbf{s}$$ and for any two arbitrary points A and B in the field, where $\textbf{AB}$ is along $\textbf{E}$ (similar to your diagram) you get: $$V_B - V_A = -\textbf{E}.∆\textbf{s} < 0$$ $$V_B < V_A$$

Showing that the potential decreases in the direction of the electric field.

Hope this helps.