Gauss Law Problem to determine F/A

AI Thread Summary
The discussion revolves around calculating the force per unit area (F/A) on three parallel charged sheets using Gauss's Law. The sheets have surface charge densities of 7.9 nC/m², -3.3 nC/m², and 5.4 nC/m², respectively. Initial attempts to derive the electric field and force per unit area included equations like E = σ/2ε₀ and F = q'E, but the results were deemed incorrect. Participants are seeking clarification on the correct application of these principles to find the forces acting on each sheet. The conversation emphasizes the need for accurate calculations and understanding of the relationships between charge density and electric fields.
max190us
Messages
2
Reaction score
0
Three large but thin charged sheets are parallel to each other as shown in the figure . Sheet I has a total surface charge density of 7.9 nC/m^2, sheet II a charge of -3.3 nC/m^2, and sheet III a charge of 5.4 nC/m^2.

Figure:

----------------(I)
----------------(II)----------------(III)

Estimate the magnitude of the force per unit area on sheet I, II, and III, N/m^2.

**I do not know how to start**

Attempt: E=F/Q; Int(E) dA=Q/epsilon

2. E=F/Q; Int(E) dA=Q/epsilon; \sigma=Q/A
3. I combined the above eq and got this:
\sigma*(\sigma)/(2epsilon)
 
Physics news on Phys.org


The attempt I have done:

E = σ/2εo

The force per unit area on sheet I due to sheet II is:

F = q’E = σ’AE => F/A = σ’E = σ’(σ/2εo) = 3.3x10^-9(7.9x10^-9)/2*8.85x10^-12

F/A = 1.47x10^-6 N/m²

or
FI/A = -σσ’/εo= (3.26x10^-6),

FII/A = -σσ’/εo + σσ”/εo

FIII/A = σ’σ”/εo


---IT IS NOT RIGHT! HELP ME!
 
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