Answer :
Answer:
V(h) = Eh
Explanation:
I will assume that the capacitor is a parallel-plate capacitor.
By Gauss' Law, electric field inside the capacitor is
[tex]E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}[/tex]
The relation between electric field and potential is
[tex]V_{ab} = -\int\limits^b_a {\vec{E}(h)} \, d\vec{h} = \int\limits^h_0 {\frac{Q}{\epsilon_0 A}} \, dh \\V(h) - V(0) = V(h) - 0 = Eh\\V(h) = Eh[/tex]
The important thing in this question is that the electric field inside the parallel plate is constant. So, the potential is also constant and proportional to the distance, h.