0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ why doesn;t anyone know about Wikipedia. en.wikipedia.org/wiki/Dielectric $\endgroup$ – JMLCarter Jan 13 '17 at 11:50
  • $\begingroup$ Why do you think surface charge density should not change? "By my knowledge" is not a reason. $\endgroup$ – sammy gerbil Jan 13 '17 at 19:17
1
$\begingroup$

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,

$\endgroup$
0
$\begingroup$

Recall that $Q = CV$. Here, you are connected to the 1500 V source, so that part of the equation cannot change.

What about $C$, the capacitance? Well, the formula for capacitance between two parallel plates is $$ C = \frac{\epsilon A}{d} = \frac{\kappa\epsilon_0A}{d}.$$ If you insert a dielectric into the middle, then you are increasing $\kappa$, which means you are increasing $C$, because you are holding everything constant. Therefore, in our equation $Q = CV$, $Q$ has to increase as well. So you increase the amount of charge on the plate, and thus the surface charge density.

Conceptually speaking, the dielectric polarizes in response to the electric field from the plates. The polarization is such that the overall net electric field is reduced, so there is less opposition to building up more charge.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.