Respuesta :
Answer:
[tex]n=4.31\times 10^{11}[/tex]
Explanation:
It is given that,
Electric field due to the spark, [tex]E=3\times 10^6\ N/C[/tex]
Dimensions of the parallel-plate capacitor is 5.1 cm × 5.1 cm.
The area of the parallel plate, [tex]A=26.01\ cm^2=0.002601\ m^2[/tex]
The electric field due to the parallel plate capacitor is given by :
[tex]E=\dfrac{q}{A\epsilon_o}[/tex]
[tex]q=EA\epsilon_o[/tex]
[tex]q=3\times 10^6\times 0.002601\times 8.85\times 10^{-12}[/tex]
[tex]q=6.90\times 10^{-8}\ C[/tex]
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 :
[tex]q=ne[/tex]
[tex]n=\dfrac{q}{e}[/tex]
[tex]n=\dfrac{6.90\times 10^{-8}}{1.6\times 10^{-19}}[/tex]
[tex]n=4.31\times 10^{11}[/tex]
So, the number of electrons must be transferred from one plate to the other is [tex]4.31\times 10^{11}[/tex]. Hence, this is the required solution.
Answer:
number of electron transferred is [tex]4.31\times 10^{11}[/tex] electron
Explanation:
Given data:
Area of the parallel plate capacitor is
A = 0.051 × 0.051 m
Electric field for parallel capacitor is
[tex]E = \frac{\sigma}{\epsilon}[/tex]
[tex]E = \frac{Q}{A \epsilon}[/tex]
[tex]Q = EA\epsilon[/tex]
[tex] = 3\times 10^{6} \times 0.0026 (8.854\times 10^{-12}})
[tex]= 6.90 \times 10^{-8}[/tex]
[tex]=69 \times 10^{-9}[/tex]
[tex]C = 69 nC[/tex]
Q = ne
[tex]n = \frac{69 \times 10^{-9}}{1.6\times 10^{-19}}[/tex]
[tex]n = 4.31\times 10^{11}[/tex]
therefore number of electron transferred is [tex]4.31\times 10^{11}[/tex] electron
