"Dry air will break down if the electric field exceeds about 3.0×106V/m"
Q) What amount of charge can be placed on a capacitor if the area of each plate is 9.2 cm2 ?
"Dry air will break down if the electric field exceeds about 3.0×106V/m" Q) What amount of...
Dry air will break down if the electric field exceeds about3.0 × 106 V/m. What amount of charge can be placedon a capacitor if the area of each plate is 3.1 cm2? nC I tried using this formula rearranged for Q : Q = Eε0A = (3.0 × 106 V/m)(8.85 x10-12 Nm2/C2)(.031m2) = 8.2305 x 10-7C = 8.2405e-7nC However, this is not right. Can anyone tell me what I amdoing wrong?
The dielectric strength of air, E = 3.0×106 V/m, is the maximum field that air can withstand before it breaks down and becomes conducting. How much charge can be placed on a spherical conductor with a 8.0- cm radius before the field at its surface exceeds the breakdown strength of the air? The answer to this section is 2.14×10-6 C. My question: What would be the electric potential at the surface of this conductor?
Suppose you build an air-filled parallel-plate capacitor with a radius equal to 3.1 m and a plate separation of 1.0 mm. The largest electric field that air can sustain before its insulating properties break down and it begins conducting electricity is 3.0 MV/m. What is the maximum charge that you will be able to store in the capacitor?
A parallel-plate air-filled capacitor having area 31 cm2 and plate spacing 3.0 mm is charged to a potential difference of 400 V. Find the following values. (a) the capacitance pF (b) the magnitude of the charge on each plate nC (c) the stored energy μJ (d) the electric field between the plates V/m (e) the energy density between the plates J/m3
If the electric field of a beam of light exceeds the breakdown value for air, the beam does not propagate through the air. The maximum electric field that can propagate through the air is 3.0 x 106 V/m. (a) What is the maximum amplitude of the magnetic field for this light wave. {0.01 T} Calculate the intensity of this wave. {1.19 x 109 W/m2} Describe what possibly happens when the wave intensity is higher than this value.
A 3.0 cm × 3.0 cm parallel-plate capacitor has a 2.0 mm spacing. The electric field strength inside the capacitor is 1.4×105 V/m . What is the potential difference across the capacitor? How much charge is on each plate? Ive been trying but i get nothing but wrong answers. My answers for A were 414.3 and 420 but its wrong. Can you please be detail so I can understand
In the figure below, two parallel-plate capacitors (with air
between the plates) are connected to a battery. Capacitor 1 has a
plate area of 1.5 cm2 and an electric field (between its
plates) of magnitude 2000 V/m.
Capacitor 2 has a plate area of 0.70 cm2 and an electric
field (between its plates) of magnitude 1300 V/m. What is the total charge on the
two capacitors?
The answer is not q = εo
E1A1 + εo
E2A2 = 8.85 e-12 (...
Consider the Earth and a cloud layer 500 m above the planet to be the plates of a parallel-plate capacitor. (a) If the cloud layer has an area of 3.0 km2 = 3000000 m2, what is the capacitance? F (b) If an electric field strength greater than 3.0 ✕ 106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold? C (Please show work)
6. Consider the Earth and a cloud layer 700 m above the planet to be the plates of a parallel - plate capacitor. (a) If the cloud layer has an area of 3.0 km^2 = 3000000 m^2, what is the capacitance?(b) If an electric field strength greater than 3.0 . 10^6 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold?7. Given a 2.75 muF capacitor, a 3.75 muF capacitor,...
QUESTION 6 What happens to the electric field in the capacitor as the charge, Q, stored in it increases? The field increases as Q. The field decreases as 1/Q. The field increases as Q2. The field decreases as 17Q2. QUESTION 7 A parallel plate capacitor Creates an electric field by pulling electrons off of atoms in the air Creates a uniform electric field between two oppositely charged plates Creates a uniform electric field between two plates with the same sign...