Electric field between parallel plates

For two large parallel conducting plates separated by distance $d = 10\,\text{cm}$ with potential difference $V$, the electric field between them is uniform:

$E = \frac{V}{d} = \frac{V}{10\,\text{cm}}$

Key idea: potential depends only on horizontal separation

In a uniform electric field between parallel plates, equipotential surfaces are planes parallel to the plates. Hence potential changes only along the direction perpendicular to the plates.

So if points $A$ and $B$ are located at different positions between the plates, then

$\Delta V_{AB} = E \times (\text{horizontal separation between }A\text{ and }B)$

From the figure

From the given geometry, the horizontal separation between $A$ and $B$ is $4\,\text{cm}$.

Therefore,

$V_A - V_B = E \times 4\,\text{cm}$

Using $E = \dfrac{V}{10\,\text{cm}}$,

$V_A - V_B = \frac{V}{10} \times 4 = \frac{2V}{5}$

Final answer

$\boxed{\frac{2}{5}V}$

Hence, the correct option is D.