Capacitors: find the dielectric constant

In summary: The correct expression is:K= ((8.85E-12)(7.8E-6 F))/(1.5(1.3E-5)) = 0.0378In summary, the dielectric constant of the dielectric in a parallel plate capacitor with a capacitance of 7.8 μF, an area of 1.5 m^2, and a separation of 1.3 x 10-5 m is 0.0378.
  • #1
Alice7979
36
2

Homework Statement


A parallel plate capacitor has a capacitance of 7.8 μF when filled with a dielectric. The area of each plate is 1.5 m^2 and the separation between the plates is 1.3 x 10-5 m. What is the dielectric constant of the dielectric?

Homework Equations


C= (K(8.85E-12)A)/d

The Attempt at a Solution


K= (1.5(7.8E-6 F))/((8.85E-12)(1.3E-5)
=1.02E11
 
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  • #2
Alice7979 said:

Homework Statement


A parallel plate capacitor has a capacitance of 7.8 μF when filled with a dielectric. The area of each plate is 1.5 m^2 and the separation between the plates is 1.3 x 10-5 m. What is the dielectric constant of the dielectric?

Homework Equations


C= (K(8.85E-12)A)/d

The Attempt at a Solution


K= (1.5(7.8E-6 F))/((8.85E-12)(1.3E-5)
=1.02E11
Your last expression is unclear because the parentheses do not work (I am assuming you forgot a closing parenthesis at the very end). Double check your algebra, it looks like you end up multiplying the capacitance by the area in finding K, which cannot be right.
 
  • #3
Alice7979 said:

Homework Statement


A parallel plate capacitor has a capacitance of 7.8 μF when filled with a dielectric. The area of each plate is 1.5 m^2 and the separation between the plates is 1.3 x 10-5 m. What is the dielectric constant of the dielectric?

Homework Equations


C= (K(8.85E-12)A)/d

The Attempt at a Solution


K= (1.5(7.8E-6 F))/((8.85E-12)(1.3E-5)
=1.02E11
I had the area and length switched, thank you
 
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