Determine the dielectric constant

[jamba88]

Homework Statement

Two flat metal circles of radius r=15 mm are brought a distance d=8.0 mm apart to form a parallel plate capacitor. The surface charge density of one plate is measured to be +1.0*10^-3 microCouloubs/mm^2. The potential difference across the capacitor is measured to be .12 Volts. There is a dielectric between the plates. What is the dielectric constant.

Homework Equations

C=K*(E0)*[A/d]
A=(pi)r^2
Q= CV... maybe

The Attempt at a Solution

I plugged what I had into the equation, but I don't know how to find the capacitance using the surface charge density.
I have
C= K(8.85*10^-12)*[pi*(.015^2)/.008]

 

[rock.freak667]

Find the area of the plate(mm²), then multiply that by the density to give charge.

 

[nlightner]

= K*(5.89*10^-8)
But I'm not sure where to go from here.

Dear student,

Thank you for your question. To determine the dielectric constant, we need to find the capacitance of the parallel plate capacitor. The equation you used, C=K*(E0)*[A/d], is correct. However, to find the capacitance, we need to use the value of the surface charge density (σ) in the equation Q=σA, where Q is the charge on one of the plates and A is the area of the plate.

So, from this equation, we can write Q = σ*(pi*r^2) = (1.0*10^-3 * 10^-6)*(pi*0.015^2) = 7.07*10^-12 C.

Now, we can plug this value of Q into the equation Q=CV to get the capacitance, C = Q/V = (7.07*10^-12 C)/0.12 V = 5.89*10^-11 F.

Finally, we can plug this value of capacitance into the original equation C=K*(E0)*[A/d] and solve for K.

K = C/[(E0)*A/d] = (5.89*10^-11 F)/[(8.85*10^-12 F/m)*(pi*0.015^2 m^2)/0.008 m] = 2.26.

Therefore, the dielectric constant for the material between the plates is approximately 2.26.

I hope this helps. Keep up the good work in your studies!

Best regards,