How Do You Calculate the Charge on the Inner Sphere Using Gauss's Law?

AI Thread Summary
To calculate the charge on the inner sphere using Gauss's Law, the radial electric field of 7.366e+04 N/C at a distance of 3.125 cm is utilized. The electric flux through the Gaussian surface is calculated using the area of the sphere, yielding a value of 2.89e+07 Nm^2/C. Rearranging Gauss's Law allows for the determination of the charge enclosed, resulting in a charge of 2.89e+04 C, which converts to 2.89e+13 nC. It is crucial to ensure consistent units throughout the calculations. The final charge on the inner sphere is 2.89e+13 nC.
frogstomp_theory
Two charged concentric metal spheres have radii of 3.000 cm and 3.500 cm. Calculate the charge in nC on the inner sphere if the radial electric field at a radial distance 3.125 cm is 7.366e+04 N/C.

I tried using Gausses law of EA = Q/e0 then after getting
Q I would take the average charge per area. Then I would simply enter into the equation Q=o(greek letter)A the area of the inner sphere, but that's wrong:frown: any suggestions?

I've been having trouble with magnetism so any help is appreciated, thanks.
 
Physics news on Phys.org
Hi, and welcome to PF.

Originally posted by frogstomp_theory
I tried using Gausses law of EA = Q/e0 then after getting
Q I would take the average charge per area. Then I would simply enter into the equation Q=o(greek letter)A the area of the inner sphere[/color], but that's wrong:frown: any suggestions?

I've highlighted your mistake[/color] above. You should be using the area of the Gaussian surface, not the inner sphere.

I've been having trouble with magnetism so any help is appreciated, thanks.

Actually, this is electrostatics and has nothing to do with magnetism.
 


Based on the given information, we can use Gauss's law to calculate the charge on the inner sphere. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). In this case, the inner sphere acts as the closed surface, with the radial distance of 3.125 cm being the radius of the surface.

First, we need to calculate the electric flux through the surface. The electric field is given as 7.366e+04 N/C, and the area of the inner sphere can be calculated using the formula for the surface area of a sphere (4πr^2). Plugging in the values, we get the electric flux to be 2.89e+07 Nm^2/C.

Next, we can rearrange Gauss's law to solve for the charge enclosed by the inner sphere. This gives us Q = ε0 * electric flux. Plugging in the values, we get the charge on the inner sphere to be 2.89e+04 C. However, the given unit for charge is in nanocoulombs (nC), so we need to convert it. 1 C = 1e+9 nC, so the charge on the inner sphere is 2.89e+13 nC.

In conclusion, the charge on the inner sphere is 2.89e+13 nC. It is important to double check the units and make sure they are consistent throughout the calculation. Hope this helps!
 
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