Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is :

Between parallel plates, electric field is uniform.

If plate separation is 10cm and potential difference is V, field is

$$E=\frac{V}{10}​(volts/cm)$$

We need potential difference between A and B.

From figure:

AC=3cm

AB=5 cm

Triangle ACB is right angled, so

$$CB=\sqrt{\ 5^2-3^2}$$

Now electric field is horizontal (between vertical plates), so only horizontal separation matters for potential difference.

Vertical displacement does not affect potential.

Thus only CB=4 cm contributes.

So

$$V_{AB}=E(4)$$

$$=\frac{V}{10}\times4=$$

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