Increase in Charge on a Parallel Plate Capacitor

[Drakkith]

Homework Statement

The charge on the 3.00 cm2 area plates of an air-filled parallel plate capacitor is such that the electric field is at the breakdown value. By what factor will the maximum charge on the plates increase when bakelite is inserted between the plates? (The dielectric strength of air is 3.00 ✕ 10⁶ V/m and that of bakelite is 2.40 ✕ 10⁷ V/m. The dielectric constant of bakelite is 4.90.)

Homework Equations

C=Q/V
C=kC₀
C₀=ε₀A/d
V=Ed

The Attempt at a Solution

I'm not sure what to do here. I don't even have a strategy to tackle this problem. So far I've tried manipulating all the equations I can find that have to do with capacitors, but it hasn't gotten me anywhere.

 

[Drakkith]

Okay, I managed to solve it.

If Q=CV, then Q₀=C₀V₀ and Q₁ = C₁V₁.
Since V₀ is the breakdown voltage of air, Q₀ = 3x10⁶C₀.
C₁=4.9C₀, so Q₁ = 4.9C₀V₁. But V₁ is the dielectric breakdown voltage.
So Q₁=(4.9)(2.4x10⁷)C₀.
The ratio of increase is Q₁/Q₀, so that becomes 1.176x10⁸C₀/3x10⁶C₀.
C₀ cancels out of the top and bottom and we get Q₁/Q₀ = 39.2.

 

[gneill]

That works. You could also have approached it via the electric field strength given the charge density on the plates, $E = \frac{\sigma}{\epsilon_o}$, as the charge is proportional to the charge density and the plate area is fixed. Rearrange as $\sigma = \epsilon_o E$ and remember that the dielectric constant multiplies $\epsilon_o$ for the bakelite. Form a ratio for the two materials and you're done.

 

[Drakkith]

Thanks gneill!