The capacitor shown in the figure has parallel plates, with each plate having an area $A$. The thickness of the dielectric materials are $d_{1}$ and $d_{2}$ and their relative permittivities are $\varepsilon_{1}$ and $\varepsilon_{2}$, respectively. Assume that the fringing field effects are negligible and $\varepsilon_{0}$ is the permittivity of free space.
If $\mathrm{d}_{1}$ is decreased by $\delta \mathrm{d}_{1}$, the resultant capacitance becomes
$\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}_{1}-\delta \mathrm{d}_{1}+\frac{\mathrm{d}_{2}}{\varepsilon_{2}}}$
$\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}_{2}+\frac{\mathrm{d}_{1}}{\varepsilon_{2}}}$
$\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}_{2}-\delta \mathrm{d}_{2}+\frac{\mathrm{d}_{1}}{\varepsilon_{2}}}$
$\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}_{1}+\delta \mathrm{d}_{1}+\frac{\mathrm{d}_{2}}{\varepsilon_{2}}}$