Capacitor with radius finding the energy density

AI Thread Summary
A capacitor is created from two concentric spherical shells with a potential difference of 100 V applied. To find the energy density at specific points, the correct formulas for a spherical capacitor must be used, as the standard parallel plate capacitor equations are not applicable. The energy density can be calculated using the formula (1/2) ε₀ E², where E is the electric field. The electric field for a spherical capacitor can be determined using the relationship E = V/d, where d is the distance from the center. Accurate calculations require careful application of these principles and proper unit handling.
sonrie
Messages
35
Reaction score
0
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.5 cm , and the outer sphere has radius 16.5 cm . A potential difference of 100 V is applied to the capacitor.



What is the energy density at r = 12.6 cm , just outside the inner sphere?

What is the energy density at r = 16.4 cm , just inside the outer sphere?

Equations:
U= 1/2 CV^2/Ad
U= 1/2 Eo E^2
C= EoA/d
V= Ed

C= 8.85*10^-12 *pi * .126^2/.04 =1.11*10^-11
so next i solved for U, U= .5 *1.11*10^-11 *100^2/pi*.126^2*.04=2.78*10^-5 which is not correct, Help PLease!
 
Physics news on Phys.org
Hi sonrie,

I believe you are using a wrong formula here. The expression

<br /> C=\epsilon_0 \frac{A}{d}<br />

applies to a parallel plate capacitor. The spherical capacitor has a different formula.

The formula energy density = (1/2) CV^2/(Ad) is also normally used for the constant field of a parallel plate capacitor. The other one (energy density = (1/2) \epsilon_0 E^2) applies to any capacitor problem.

(You might also find that the formula C=Q/V is helpful.)
 
Thanks!
 
How do I find E, which I need to find the energy density? Then what do I do?
 
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