Electric Field of a charged sphere with cylindrical gaussian surface

[Molderish]

So the problem statement is:

A conducting solid sphere (R = 0.167 m, q = 6.63·10–6 C) is shown in the figure. Using Gauss’s Law and two different Gaussian surfaces, determine the electric field (magnitude and direction) at point A, which is 0.00000100 m outside the conducting sphere. (Hint: One Gaussian surface is a sphere, and the other is a small right cylinder.)

Now with the spherical gaussian surface I've got no doubt..

The problem with the cylindrical surface is that the electric field it's not constant ,is it?
in some parts the "area vector" is parallel to the vector of electric field and in some part there's an angle.

So I'm not sure if i have to find a function of the electric field with respect to the radius of the sphere or how to get ride of this problem.

What I've first tried was to "assume the electric field" was constant so integrating dA what I've got:
since the height of cylinder is h=2r (of the sphere).
and there's flux all over the cyinder: E(2pi*h+2pi*r^2)=Q/ε...
E(4pi*r+2pi*r^2)=Q/ε... and so E=Q/(4pi*r+2pi*r^2)ε... but I'm not sure if a i could assume that...

 

[BvU]

Reconsider your choice of small cylinder. Guiding principle: don't look for more difficult things to solve but try to find something even easier. Your clues:
1. the kind of answer to the first part of the exercise,
2. conducting sphere,
3. 0.00000100 m is very, very close to the surface

 

[Molderish]

i could reconsider however it's something I'm being asked for. :(.

yep i know the answer must be E=Q/(4pi*r^2)ε.

and i know also the charge on the sphere it's all over the "surface area" and that the distance from the center to the point would be R+0.00000100 m.

Once i got the Electric field ussing a cylindrical gaussian surface it's just substitution. the problem is i need but I'm not sure how to reach the same result mathematically using that surface :/

 

[BvU]

1. the kind of answer to the first part of the exercise,
2. conducting sphere,
3. 0.00000100 m is very, very close to the surface

2. so all the charge is at the surface, and uniformly distributed.
3. so close to the surface that field lines over there are parallel
So a really puny cylinder standing upright on an almost flat surface would be a nice idea. floor just under the surface, top 0.00000100 m or thereabouts above...

 

[paranormal]

I would suggest approaching this problem by considering the geometry and symmetry of the situation. The electric field of a charged sphere will always be radial and symmetric, meaning it will have the same magnitude and direction at all points on a given radius from the center of the sphere. Therefore, the electric field at point A, which is outside the sphere, will also be radial and symmetric.

To use Gauss's Law, we need to choose a Gaussian surface that takes advantage of this symmetry. For the spherical Gaussian surface, we can easily calculate the electric field at point A by using the equation E = Q/4πε0r^2, where Q is the charge enclosed by the surface and r is the distance from the center of the sphere to point A.

For the cylindrical Gaussian surface, we can still use Gauss's Law by considering the symmetry of the electric field. The electric field will be constant along the curved surface of the cylinder, but it will vary along the flat ends. We can use the equation E = Q/2πε0L, where Q is the charge enclosed by the surface and L is the length of the cylinder, to calculate the electric field along the curved part of the cylinder. Then, we can use the component of the electric field that is perpendicular to the flat ends to calculate the total electric field at point A.

In conclusion, the key to solving this problem is to consider the symmetry of the electric field and choose a Gaussian surface that takes advantage of this symmetry. By doing so, we can easily calculate the electric field at point A using Gauss's Law.