Induced charge density at the interface of dielectric slabs

[Kaguro]

Homework Statement

In a parallel plate capacitor the distance between the plates is 10cm. 2 dielectric slabs of thickness 5 cm each and dielectric constants 2 and 4 respectively are inserted between the plates. A potential difference of 10V is applied between the plates. Find the net bound charge density at the interface of the two dielectrics.

Relevant Equations

E0=V/d
E=E0/k
Gauss' Law

E0=V/d = 100/0.1 =1000v/m

In slab 1, E1=E0/k1=500v/m
In slab 2, E2=E0/k2=250v/m

Applying Gauss' Law to a box surface surrounding the interface with area equal to the plates we have

(-E1+E2)*A = Q/epsilon_naught

So charge density sigma = -250 epsilon_naught

But answer given is (-2000/3)*epsilon_naught

What did i do wrong?

 

[Kaguro]

Sorry, the applied potential difference is 100v not 10v.

 

[TSny]

E0=V/d = 100/0.1 =1000v/m

In slab 1, E1=E0/k1=500v/m
In slab 2, E2=E0/k2=250v/m

You want the potential difference to be 100 V when the dielectric slabs are in place. If you use your values of E1 and E2, do you get 100 V between the plates?

 

[Kaguro]

You want the potential difference to be 100 V when the dielectric slabs are in place. If you use your values of E1 and E2, do you get 100 V between the plates?

Ok!
I understand now!

Let E1 =2x and so E2=x

E1*(d/2) +E2*(d/2) = 100
So 3x*0.05=100
x=2000/3 v/m

Now
Using Gauss' Law
(-4000/3 + 2000/3)*A= Q/epsilon_nought

So sigma= -2000/3 * epsilon_nought

Thanks very much!
I love PF!