Answer:
[tex]V = \frac{V_0x}{d}[/tex]
Explanation:
Since the field lines are parallel and the electric field is uniform between two parallel plates, a test charge would experience the same force of attraction or repulsion no matter where it is located in the field,
I attached an image that could help to understand the representation of the field. The formula used to calculate it is given by,
[tex]E= -\frac{\Delta V}{x}[/tex] (1)
If we want to consider the change in Voltage with respect to the position then it would be,
[tex]E= -\frac{dV}{dx}[/tex]
According to the information provided, the potential is [tex]V_0[/tex] and there is a distance d, therefore
[tex]E= -\frac{V_0}{d}[/tex] (2)
Taking equation (1) we can clear V, to what we have,
[tex]\frac{dV}{dx} = -E[/tex]
[tex]dV = -Edx[/tex]
Integrating,
[tex]V= - \int Edx[/tex]
Substituting (2)
[tex]V = -\int \frac{V_0}{d} dx[/tex]
[tex]V = \frac{V_0x}{d}[/tex]
Where x is the height from the grounded plate.
