Electric Potential at Cylinder Center: Calculating Error

AI Thread Summary
The discussion focuses on calculating the electric potential at the center of a hollow cylindrical shell with a uniform charge distribution. The initial calculation contains errors, particularly in the expression for dq, which should be dq = eta * dL, and the integral limits that must encompass the entire length of the cylinder. Additionally, the height of the cylinder must be included in the denominator of the potential equation. The direction of the electric field at the center is also crucial for determining the potential. Careful consideration of geometry, units, and constants is recommended to ensure accuracy in the calculations.
brichugh
Messages
1
Reaction score
0
A hollow cylindrical shell of length L and radius R has charge Q uniformly distributed along its length. What is the electric potential at the center of the cylinder?

I use eta = Q/A = Q/2piRL

dq = (eta)*dA = Q/(2piRL)*2piRdL

dV = 1 / 4 pi epsilon 0 * dq/r where r = sqrt(L^2+R^2)

V = Q/(4*pi*epsilon_0)*int(dL/(sqrt(L^2+R^2)))

This seems to be incorrect can someone tell me where I have gone wrong?

Thanks in advance.
 
Physics news on Phys.org
You are missing an H = height of the cylinder in the denominator of the equation.


Also limits of the integral (-h/2 -> h/2) if your referance point is the middle of the cylinder.
 
Last edited:


There are a few potential sources of error in your calculation. Firstly, the expression for dq should be dq = eta * dL, since dA is equal to R * dL for a cylindrical shell. Secondly, the integral should be taken over the entire length of the cylinder, not just a small portion of it. Additionally, it is important to consider the direction of the electric field at the center of the cylinder, since the potential will depend on the direction of the field. It may be helpful to draw a diagram and carefully consider the geometry and direction of the electric field lines. I would also recommend double checking the units and constants used in the calculation to ensure they are consistent. Overall, it is always a good idea to double check your work and carefully consider the physical meaning of the equations being used.
 
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