# Dielectrics and guass' law

1. Sep 18, 2007

### roro0505

the answer i keep getting is wrong ... help?

1. The problem statement, all variables and given/known data

Two parallel plates of area 110 cm2 are given charges of equal magnitudes 9.6 × 10-7 C but opposite signs. The electric field within the dielectric material filling the space between the plates is 4.2 × 106 V/m.

(a) Calculate the dielectric constant of the material.
(b) Determine the magnitude of the charge induced on each dielectric surface.

2. Relevant equations

k=q/(E*A*epsilon_0)

q'=q(1-(1/k))

3. The attempt at a solution

k=q/(E*A*epsilon_0)
=9.6E-7/(4.2E6*110E-4*8.85E-12)
=2.3

q'=q(1-(1/k))
=9.6E-7(1-(1/2.3))
=5.4E-7

2. Sep 18, 2007

### Andrew Mason

It is not clear but I suspect that these are non-conducting plates, in which case Gauss' law is:

$$\oint E dA = E2A = \frac{Q}{k\epsilon_0}$$

AM