Capacitance:

$\overline{){\mathbf{C}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}}$

k is the dielectric constant, A is the area and d is the plate separation.

k = Cd/ε_{0}A

(1 m)^{2} = (100cm)^{2}

k_{1} = (d/ε_{0}) × (2 × 10^{-9})/(4 × 10^{-4}) = (5.0 × 10^{-6})(d/ε_{0})

Six parallel-plate capacitors of identical plate separation have different plate areas ,different capacitances ,and different dielectrics filling the space between the plates.

Part A. Rank the following capacitors on the basis of the dielectric constant of the material between the plates.

Rank from largest to smallest.

1. A = 4 cm^{2} C = 2 nF

2. A = 1 cm^{2} C = 1 nF

3. A = 2 cm^{2} C = 8 nF

4. A = 8 cm^{2} C = 2 nF

5. A = 4 cm^{2} C = 1 nF

6. A = 2 cm^{2} C = 4 nF

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Capacitors & Capacitance concept. You can view video lessons to learn Capacitors & Capacitance. Or if you need more Capacitors & Capacitance practice, you can also practice Capacitors & Capacitance practice problems.