Electrostatics - sphere electric field

AI Thread Summary
The discussion focuses on calculating the electric field intensity at two distances from the center of two concentric spheres with given radii and charges. At a distance of 5 cm, which is inside both spheres, the electric field is zero due to the properties of conductors. For the distance of 14 cm, which is outside the spheres, the electric field can be calculated using the formula E = KQ/R^2, treating the spheres as point charges. An alternative method suggested is using Gauss's law to derive the electric field. The conversation emphasizes the need for clarity on the formulas and methods to solve the problem.
Agonche
Messages
1
Reaction score
0

Homework Statement


Two concentric spheres with radius r1=6cm and r2=10cm are charged with electric charges Q1=-1nC and Q2=0.5nC. Find the intensity of the electric field on distance 5cm and 14 cm from the center of the sphere.

I think they look something like this

tan19s.jpg


Please if anyone knows the solution. write it down.
I just need the start .. the formula.. i can do calculations by myself.

Homework Equations





The Attempt at a Solution



i may be wrong but here's my try
since the first distance d1=5cm
and the sphere sizes are larger so
d1<r1
d1<r2
E1=0
E2=0

there is no electric field inside sferes

what about the distance 14cm
 
Last edited:
Physics news on Phys.org
Agonche said:

Homework Statement


Two concentric spheres with radius r1=6cm and r2=10cm are charged with electric charges Q1=-1nC and Q2=0.5nC. Find the intensity of the electric field on distance 5cm and 14 cm from the center of the sphere.

I think they look something like this

tan19s.jpg


Please if anyone knows the solution. write it down.
I just need the start .. the formula.. i can do calculations by myself.

Homework Equations


The Attempt at a Solution



i may be wrong but here's my try
since the first distance d1=5cm
and the sphere sizes are larger so
d1<r1
d1<r2
E1=0
E2=0

there is no electric field inside sferes

what about the distance 14cm

For E field at 14cm, since this point is outside of the sphere,take the sphere as a point charge and calculate the E-field using E=KQ/R^2 (E-field of point charge)

an alternative view point will be to use gauss law, EA=Q(enclosed)/(epsilon nought)
manipulate the equation and solve for E.
 
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