If empty (filled with vacuum) parallel plate capacitor has two plates set to be $ d=0.0012m $ apart and connected to $ 1500 V $ voltage source, then surface charge density should be:
$$ \sigma = \frac{\varepsilon_0 U}{d} \approx 1.107 C/m^2 $$
Now we insert dielectric with width $ w=0.0006m $ so that it touches one of the plates. By my knowledge that shouldn't change surface charge density, however solutions to this problem assume new surface charge density to be $\sigma_{new} \approx 1.72 C/m^2$.
How is that possible?
Best Answer
Introduction of dielectric will cause capacitance to change which in turn cause the charge densities to change. In other words, the charge on inner side of any single plate will be redistributed to maintain the effect of dielectric.
$$ C_{old} = \epsilon_0 \, \frac {Area} {distance} $$
$$ C_{new} = \epsilon(r) \,\epsilon_0 \, \frac {Area} {distance}$$
This change in surface charge densities is only due to change in capacitance,