Capacitance and maximum charge HELP

AI Thread Summary
The discussion focuses on calculating the capacitance of a parallel-plate capacitor modeled as a cloud and the ground, with the cloud at 620 meters altitude and a surface area of 0.16 km². The formula used is C=E0 A/D, where E0 is the permittivity of free space. Participants are seeking assistance with the calculations, as initial attempts did not yield the correct capacitance in nanofarads (nF) or the maximum charge before lightning occurs. The thread emphasizes the need for accurate conversions and calculations to arrive at the final answers. Accurate capacitance and charge values are crucial for understanding the lightning phenomenon modeled in this scenario.
seant
Messages
1
Reaction score
0
As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an altitude of 620 m to be the other plate. Assume the surface area of the cloud to be the same as the area of a square that is 0.40 km on a side.
(a) What is the capacitance of this capacitor?
nF
(b) How much charge can the cloud hold before the dielectric strength of the air is exceeded and a spark (lightning) results?
C

I used C=E0 A/D (8.85*10^-12 C^2/N*m^2)(.16km^2/620m) and conversions also, but to no avail can someone please help
 
Physics news on Phys.org
Hi seant,

seant said:
As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an altitude of 620 m to be the other plate. Assume the surface area of the cloud to be the same as the area of a square that is 0.40 km on a side.
(a) What is the capacitance of this capacitor?
nF
(b) How much charge can the cloud hold before the dielectric strength of the air is exceeded and a spark (lightning) results?
C

I used C=E0 A/D (8.85*10^-12 C^2/N*m^2)(.16km^2/620m) and conversions also, but to no avail can someone please help

What number did you get as your final answer, in nF?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top