Respuesta :
Electric field between the plates of parallel plate capacitor is given as
[tex]E = \frac{Q}{A\epsilon_0}[/tex]
here area of plates of capacitor is given as
[tex]A = 0.055 * 0.055[/tex]
[tex]A = 3.025 * 10^{-3}[/tex]
also the maximum field strength is given as
[tex]E = 3 * 10^6 N/C[/tex]
now we will plug in all data to find the maximum possible charge on capacitor plates
[tex]3 * 10^6 = \frac{Q}{3.025 * 10^{-3}*8.85 * 10^{-12}}[/tex]
[tex]Q = 8.03 * 10^{-8} C[/tex]
so the maximum charge that plate will hold will be given by above
Complete Question:
Air "breaks down" when the electric field strength reaches 3×10⁶N/C, causing a spark. A parallel-plate capacitor is made from two 5.5cm × 5.5cm plates.
How many electrons must be transferred from one disk to the other to create a spark between the disks?
Answer:
5.02 x 10¹¹
Explanation:
The electric field E of a parallel-plate capacitor is given by;
E = Q / (A ε₀) --------------------------(i)
Where;
Q = the charge on the plates
A = Area of any of the plates
ε₀ = electric constant = 8.85 x 10⁻¹²F/m
From the question;
E = 3 x 10⁶N/C
A = 5.5cm x 5.5cm = 30.25cm² = 0.003025m²
Substitute these values into equation (i) as follows;
3 x 10⁶ = Q / (0.003025 x 8.85 x 10⁻¹²)
Solve for Q;
Q = 3 x 10⁶ x 0.003025 x 8.85 x 10⁻¹²
Q = 8.03 x 10⁻⁸
The charge on the capacitor is therefore, 8.03 x 10⁻⁸C
Now, to get the number of electrons that must be transferred from one disk to the other to create a spark between the disks, we use the relation;
Q = N x E ----------------------(ii)
Where;
Q = the charge = 8.03 x 10⁻⁸C [as calculated above]
N = the number of electrons
E = the charge of one electron = 1.6 x 10⁻¹⁹C [a known value]
Substitute these values into equation (ii) as follows;
8.03 x 10⁻⁸ = N x 1.6 x 10⁻¹⁹
Solve for N;
N = [8.03 x 10⁻⁸] / [1.6 x 10⁻¹⁹]
N = 5.02 x 10¹¹
Therefore, the number of electrons that must be transferred from one disk to the other to create a spark between the disks is 5.02 x 10¹³
