Question

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.40×105 V/m . When the space is filled with dielectric, the electric field is E= 2.20×105 V/m . Part A What is the charge density on each surface of the dielectric?

Answer 1

Answer:

$\sigma_i=1.06*10^{-6}C$

Explanation:

When the space is filled with dielectric, an induced opposite sign charge appears on each surface of the dielectric. This induced charge has a charge density related to the charge density on the electrodes as follows:

$\sigma_i=\sigma(1-\frac{E}{E_0})$

Where E is the eletric field with dielectric and $E_0$ is the electric filed without it. Recall that $\sigma$ is given by:

$\sigma=\epsilon_0E_0$

Replacing this and solving:

$\sigma_i=\epsilon_0E_0(1-\frac{E}{E_0})\\\sigma_i=8.85*10^{-12}\frac{C^2}{N\cdot m^2}*3.40*10^5\frac{V}{m}*(1-\frac{2.20*10^5\frac{V}{m}}{3.40*10^5\frac{V}{m}})\\\sigma_i=1.06*10^{-6}C$