Question

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular aluminum plates, each 19 cm in diameter, separated by 1.0 cm.How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3×106N/C of the field causes a spark. Express your answer with the appropriate units.

Answer 1

Answer:

$7.54\cdot 10^{-7} C$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{\epsilon_0 A}{d}$

where $\epsilon_0$ is the vacuum permittivity, A is the surface area of the plates, d their separation.

We also know the following relationship

$C=\frac{Q}{V}$

where Q is the charge stored on the capacitor and V the potential difference between the plates.

Combining the two equations,

$\frac{\epsilon_0 A}{d}=\frac{Q}{V}$

We also know that for a uniform electric field (such as the one between the plates of a parallel-plate capacitor), we have

$V= Ed$

where E is the magnitude of the electric field. Substituting into the previous equation and re-arranging it,

$\frac{\epsilon_0 A}{d}=\frac{Q}{Ed}\\Q=\frac{\epsilon_0 A E d}{d}=\epsilon_0 A E$

For the capacitor in the problem:

$A=\pi r^2 = \pi (\frac{d}{2})^2 = \pi (\frac{0.19 m}{2})^2=0.0284 m^2$ is the area of the plates

$E=3\cdot 10^6 N/C$ is the maximum electric field before a spark is produced

Solving for Q, we find the maximum charge that can be added before that occurs:

$Q=(8.85\cdot 10^{-12})(0.0284)(3\cdot 10^6)=7.54\cdot 10^{-7} C$