A composite parallel plate capacitor is made up of two different dielectric materials with different thickness ($$t_1$$ and $$t_2$$) as shown in figure. The two different dielectric materials are separated by a conducting foil $$F$$. The voltage of the conducting foil is ______ V.

Treat it as two capacitors in series with dielectric slabs separated by conducting foil F

$$\varepsilon_{r1}=3\ and\ t_1\ =\ 0.5mm$$

$$\varepsilon_{r2}=4\ and\ t_2\ =\ 1mm$$

V =100V

For dielectrics in series, same charge appears on each, so electric displacement is same, and voltage division is proportional to $$\frac{t}{\varepsilon}$$

so,

$$V_1:V_2\ =\ \frac{t_1}{\varepsilon_{r1}}:\frac{t_2}{\varepsilon_{r2}}$$

Substitute:

$$V_1:V_2=\frac{0.5}{3}:\frac{1}{4}$$

$$=\frac{1}{6}:\frac{1}{4}$$

Multiply by 12:

$$2:3$$

Total voltage is

$$V_1+V_2=100$$

So

$$V_1=40V,\quad V_2=60V$$

The conducting foil is between the two dielectric sections. Its potential (with lower plate as reference zero) equals the voltage across lower dielectric, i.e.

V=60V