Max Charge Capacity on a Spherical Body: Calculating the Limit

AI Thread Summary
The discussion revolves around calculating the maximum charge that can be stored on a spherical body based on its radius. The initial formula proposed for capacitance is C = 4πεr, leading to Q = 4πεrV, with the maximum charge expressed as Q_max = 4πεrE_0, where E_0 is the breakdown voltage. However, the contributor later realizes that knowing the electrical breakdown of air allows for a more straightforward calculation using Q = 4πεr²E(r). This correction highlights the importance of considering the electric field in determining charge capacity. The conversation concludes with an acknowledgment of the oversight and appreciation for the input received.
El Hombre Invisible
Messages
691
Reaction score
0
Me again.

Is there a formula for the maximum amount of charge may be stored on a body (assume spherical) of radius r? I assume radius is the only factor, though mass is known also. I just don't remember seeing anything for this and can't find it in all my books for the life of me.

Thanks.
 
Physics news on Phys.org
Lets try ::
Capacitance:: C=4 \pi \epsilon r
Q=CV
Q=4 \pi \epsilon r V
Now V should be less than E0 Breakdown Voltage
so max can be
Q_{max} = 4 \pi \epsilon r E_0
Well I am not too sure abt above but its a try , if u can provide some other logic it will be welcomed
 
himanshu121 said:
Lets try ::
Capacitance:: C=4 \pi \epsilon r
Q=CV
Q=4 \pi \epsilon r V
Now V should be less than E0 Breakdown Voltage
so max can be
Q_{max} = 4 \pi \epsilon r E_0
Well I am not too sure abt above but its a try , if u can provide some other logic it will be welcomed
Dammit, sorry I was being dumb. I didn't notice we were given the electrical breakdown of air in the problem, so I know the maxmum electric field, I know the radius, and from this I can easily get the charge:

Q = 4 \pi \epsilon r^2E(r)

Apologies for wasting thine time, and thanks for posting back. I was pretty tired. :zzz: Take care.
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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