Posted by **andrew** on Saturday, November 3, 2012 at 1:18am.

Two parallel plates of area 41 cm.^2 have equal but opposite charges of 2.60E-07 C. Within the dielectric material filling the space between the plates, the electric field is 5.80E+05 V./m. Find the dielectric constant of the material.

Feel like there is a huge lack of information to solve this problem, cannot find the distance separated, capacitance initially or anything of that sort

- Capacitance (tricky) -
**drwls**, Saturday, November 3, 2012 at 2:23am
C = epsilon*A/d

epsilon = epsilonzero*K

Q = C*V = epsilon*A*V/d

= epsilon*41*10^-4 m^2/d

= 2.60*10^-7 C

E = Q/(epsilon*A)= 5.8*10^5 V/m

You have two equations for the unknowns epsilon and d. You can get the dielectric constant K from the ratio

epsilon/epsilonzero

- Capacitance (tricky) -
**andrew**, Saturday, November 3, 2012 at 1:16pm
Thanks!

Our textbook did not cover dielectrics very well.

## Answer This Question

## Related Questions

- science - two parallel plates are separated by a dielectric of thickness 2mm and...
- physics - Two large circular metal plates are parallel and nearly touching, only...
- Physics - Two large, parallel, conducting plates are 20 cm apart and have ...
- Physics - How does the capacitance of a parallel plate capacitor change when its...
- physics - Two capacitors, identical except for the dielectric material between ...
- Physics - A capacitor consisting of two parallel plates of area A = 4.6 m2 and ...
- physics - Two capacitors have the same plate separation, but one has square ...
- chemistry - in a parallel plate capacitor the separation between the plates is ...
- Help!! - An air-filled capacitor consists of two parallel plates, each with an ...
- Physics - A parallel-plate capacitor has a capacitance of 2.1 µF with air ...

More Related Questions