Question

Air "breaks down" when the electric field strength reaches 3 x 10^6 N/C, causing a spark. A parallel-plate capacitor is made from two 5.1 cm x 5.1 cm plates. How many electrons must be transferred from one plate to the other to create a spark between the plates?

Answer 1

Answer:

number of electron transferred is $4.31\times 10^{11}$ electron

Explanation:

Given data:

Area of the parallel plate capacitor is

A = 0.051 × 0.051 m

Electric field for parallel capacitor is

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

$E = \frac{Q}{A \epsilon}$

$Q = EA\epsilon$

  $= 3\times 10^{6} \times 0.0026 (8.854\times 10^{-12}}) [tex]= 6.90 \times 10^{-8}$

   $=69 \times 10^{-9}$

$C = 69 nC$

  Q = ne

$n = \frac{69 \times 10^{-9}}{1.6\times 10^{-19}}$

$n = 4.31\times 10^{11}$

therefore number of electron transferred is $4.31\times 10^{11}$ electron

Answer 2

Answer:

$n=4.31\times 10^{11}$              

Explanation:

It is given that,

Electric field due to the spark, $E=3\times 10^6\ N/C$

Dimensions of the parallel-plate capacitor is 5.1 cm × 5.1 cm.

The area of the parallel plate, $A=26.01\ cm^2=0.002601\ m^2$

The electric field due to the parallel plate capacitor is given by :

$E=\dfrac{q}{A\epsilon_o}$

$q=EA\epsilon_o$

$q=3\times 10^6\times 0.002601\times 8.85\times 10^{-12}$

$q=6.90\times 10^{-8}\ C$

Let n is the number of electrons that must transferred from one plate to the other to create a spark between the plates. It can be calculated as :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{6.90\times 10^{-8}}{1.6\times 10^{-19}}$

$n=4.31\times 10^{11}$

So, the number of electrons must be transferred from one plate to the other is $4.31\times 10^{11}$. Hence, this is the required solution.