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

`\sigma_i=1.06*10^(-6)C`

Step-by-step 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-(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-(E)/(E_0))\\\sigma_i=8.85*10^(-12)(C^2)/(N\cdot m^2)*3.40*10^5(V)/(m)*(1-(2.20*10^5(V)/(m))/(3.40*10^5(V)/(m)))\\\sigma_i=1.06*10^(-6)C`