A problem involing an LC circuit

[funoras]

Homework Statement

There's an LC circuit that's made of an 31 mH inductor and a planar capacitator. The surface of the plate is 20cm^2 and the distance between the plates is 1 cm. What is the electrical permittivity of the insulator between the plates , if the peak current is 0,2mA and the peak voltage 10V
L=31mH=3.1*10^-2 H
A=20cm^2=2*10^-3 m^2
d=1cm=0,01 m
Imax=0,2mA=2*10^-4 A
Vmax=10V
ε - ?

Homework Equations

C=q/V
C=εA/d

The Attempt at a Solution

 

[berkeman]

Homework Statement

There's an LC circuit that's made of an 31 mH inductor and a planar capacitator. The surface of the plate is 20cm^2 and the distance between the plates is 1 cm. What is the electrical permittivity of the insulator between the plates , if the peak current is 0,2mA and the peak voltage 10V
L=31mH=3.1*10^-2 H
A=20cm^2=2*10^-3 m^2
d=1cm=0,01 m
Imax=0,2mA=2*10^-4 A
Vmax=10V
ε - ?

Homework Equations

C=q/V
C=εA/d

The Attempt at a Solution

Your unit conversions look okay to me.

Now, what equations can you use to relate the peak current in the LC circuit to the peak voltage? When you have peak series current, what is the voltage across the capacitor? When you have peak voltage across the capacitor, what is the current through the inductor?

 

[funoras]

maybe W=LI^2/2 and W=q^2/2C ?

 

[berkeman]

maybe W=LI^2/2 and W=q^2/2C ?

Why do you suggest these two equations? You may be on the right track, but why pick these?

Also, you should double-check the equations you wrote...

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html

.

 

[rwisz]

To solve this problem, we can use the equation for capacitance, C=q/V, where q is the charge stored on the plates and V is the voltage across the plates. We can also use the equation C=εA/d, where ε is the electrical permittivity of the insulator, A is the surface area of the plates, and d is the distance between the plates.

First, we need to find the charge stored on the plates. We can do this by using the equation q=Imax*t, where Imax is the peak current and t is the period of the oscillation. Since we are given the peak current of 0.2mA, we can calculate the period using the equation T=2π√(L/C), where L is the inductance and C is the capacitance. Plugging in the values, we get T=2π√(3.1*10^-2 H/2*10^-3 m^2)=0.00147 s. Therefore, q=0.2mA*0.00147 s=2.94*10^-7 C.

Next, we can use the equation C=q/V to find the capacitance. Plugging in the values, we get C=(2.94*10^-7 C)/(10V)=2.94*10^-8 F.

Finally, we can use the equation C=εA/d to solve for ε. Plugging in the values, we get 2.94*10^-8 F=ε*(2*10^-3 m^2)/(0.01 m). Solving for ε, we get ε=1.47*10^-7 F/m.

In conclusion, the electrical permittivity of the insulator between the plates is 1.47*10^-7 F/m.