Question

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular aluminum plates, each 21 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.

Answer 1

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

Given:

• Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
• Distance of separation between the plates, d = 1.0 cm = 0.01 m.
• Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

• $\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$.

• $\rm q$ = magnitude of the charge on each of the plate.
• $\rm A$ = surface area of each of the plate =$\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
• $\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.