Electrostatics: Find relative permitivitty

[gruba]

Homework Statement

Spherical capacitor with two linear and uniform dielectrics with relative permitivitty Ɛr1 and Ɛr2 is connected to constant voltage U. When second dielectric is removed, intensity of electric field by inner electrode is reduced by 1/3, and electric field by outer electrode is increased two times. a=1.5mm, b=6mm, c=12mm. Calculate Ɛr1 and Ɛr2.

Homework Equations

Gauss law

The Attempt at a Solution

In the first case (two dielectrics), Gauss law gives $$E_1^{(1)}=\frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2},a<r<b$$
$$E_2^{(1)}=\frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2},b<r<c$$
$$U^{(1)}=\int_a^b \frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2}\mathrm dr+ \int_b^c \frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2}\mathrm dr=\frac{Q}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{\epsilon_{r2}bc}\right)$$
In the second case (second dielectric is removed),
$$E_1^{(2)}=\frac{Q}{12\pi\epsilon_0\epsilon_{r1}r^2}$$
$$E_2^{(2)}=\frac{Q}{2\pi\epsilon_0r^2}$$
$$U^{(2)}=\frac{Q}{2\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{bc}\right)$$
$$E_1^{(2)}=\frac{1}{3}\times E_1^{(1)}\Rightarrow \frac{Q^{(2)}}{4\pi\epsilon_0\epsilon_{r1}r^2} = \frac{1}{3}\frac{Q^{(1)}}{4\pi\epsilon_0\epsilon_{r1}r^2}\Rightarrow Q^{(2)}=\frac{Q^{(1)}}{3}$$
$$E_2^{(2)}=2\times E_2^{(1)}$$
$$\frac{Q^{(2)}}{4\pi\epsilon_0 r^2} = 2\frac{Q^{(1)}}{4\pi\epsilon_0\epsilon_{r2} r^2}$$
From these equations, I get that $$\epsilon_{r2}=6$$
$$U^{(1)}=\int_a^b \frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2}\mathrm dr+ \int_b^c \frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2}\mathrm dr=\frac{Q^{(1)}}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{\epsilon_{r2}bc}\right)$$
$$U^{(2)}=\frac{Q^{(2)}}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{bc}\right)$$
$$U^{(1)}=U^{(2)}\Rightarrow \epsilon_{r1}=36017.1$$

In my book's solution $$\epsilon_{r2}=3$$ and $$\epsilon_{r1}=6$$
Could someone please check this?

 

[Ellispson]

Homework Statement

Spherical capacitor with two linear and uniform dielectrics with relative permitivitty Ɛr1 and Ɛr2 is connected to constant voltage U. When second dielectric is removed, intensity of electric field by inner electrode is reduced by 1/3, and electric field by outer electrode is increased two times. a=1.5mm, b=6mm, c=12mm. Calculate Ɛr1 and Ɛr2.

Homework Equations

Gauss law

The Attempt at a Solution

In the first case (two dielectrics), Gauss law gives $$E_1^{(1)}=\frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2},a<r<b$$
$$E_2^{(1)}=\frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2},b<r<c$$
$$U^{(1)}=\int_a^b \frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2}\mathrm dr+ \int_b^c \frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2}\mathrm dr=\frac{Q}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{\epsilon_{r2}bc}\right)$$
In the second case (second dielectric is removed),
$$E_1^{(2)}=\frac{Q}{12\pi\epsilon_0\epsilon_{r1}r^2}$$
$$E_2^{(2)}=\frac{Q}{2\pi\epsilon_0r^2}$$
$$U^{(2)}=\frac{Q}{2\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{bc}\right)$$
$$E_1^{(2)}=\frac{1}{3}\times E_1^{(1)}\Rightarrow \frac{Q^{(2)}}{4\pi\epsilon_0\epsilon_{r1}r^2} = \frac{1}{3}\frac{Q^{(1)}}{4\pi\epsilon_0\epsilon_{r1}r^2}\Rightarrow Q^{(2)}=\frac{Q^{(1)}}{3}$$
$$E_2^{(2)}=2\times E_2^{(1)}$$
$$\frac{Q^{(2)}}{4\pi\epsilon_0 r^2} = 2\frac{Q^{(1)}}{4\pi\epsilon_0\epsilon_{r2} r^2}$$
From these equations, I get that $$\epsilon_{r2}=6$$
$$U^{(1)}=\int_a^b \frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2}\mathrm dr+ \int_b^c \frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2}\mathrm dr=\frac{Q^{(1)}}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{\epsilon_{r2}bc}\right)$$
$$U^{(2)}=\frac{Q^{(2)}}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{bc}\right)$$
$$U^{(1)}=U^{(2)}\Rightarrow \epsilon_{r1}=36017.1$$

In my book's solution $$\epsilon_{r2}=3$$ and $$\epsilon_{r1}=6$$
Could someone please check this?

The question mentions that the electric field is reduced BY 1/3 and not to 1/3 as you have assumed.The rest of it seems right.