Question

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:

$9.56\cdot 10^{-7} C$

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma=\frac{Q}{A}$ is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

$A=L^2=(0.19)^2=0.036m^2$

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

$E=3\cdot 10^6 N/C$

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

$E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C$