Point Charge and Charged Sphere; Gauss' Law

[SeanFM21]

Gauss' Law has been fairly tough for me and I seem to be struggling to analyze situations properly and, specifically, decided on the net charge included in arbitrary symmetrical-shapes used for this law. Specifically, this one includes two spherical shells.

Homework Statement

A point charge q1 = -5.3 μC is located at the center of a thick conducting shell of inner radius a = 2.1 cm and outer radius b = 4.2 cm, The conducting shell has a net charge of q2 = 1.1 μC.

1) What is Ex(P), the value of the x-component of the electric field at point P, located a distance 7.2 cm along the x-axis from q1?

Homework Equations

E=Q_enc/ε_°=∫E (dot) dA
For Spherical Symmetry - E=Q_enc/4πr²ε_o

The Attempt at a Solution

Initially, I drew a (not so) arbitrary sphere of radius 7.2cm so that P is enclosed in the surface. Based off the wording of the problem, I assumed that the total Qenclosed in this arbitrary sphere is 1.1μC. Given this, I calculated 1.1μC/4∏(.072m)²ε_° which, evidently, came out to be incorrect. As I referred to earlier, it seems that I am struggling to find out what the actual enclosed charge is.

 

[Chopin]

Can you explain your reasoning for saying that the total enclosed charge is 1.1uC? How does the q1 charge factor into this problem?

 

[SeanFM21]

Can you explain your reasoning for saying that the total enclosed charge is 1.1uC? How does the q1 charge factor into this problem?

My reasoning was that despite q1 being given an explicit charge, it says that q2=1.1μC represents the net charge, which I interpret as the total charge enclosed by the whole figure. I guess I'm not completely sure how to account for q1? Once I saw that a net charge was given, I assumed that the charge in the shell of radius b=4.2cm was 6.4μC, cancelling out the q1 charge and leaving a net of +1.1μC.

 

[Chopin]

I don't think that's what the problem is saying. I would read that as saying that the total charge on the shell alone is +1.1uC, and then there's an additional charge of -5.3uC inside of the shell. The use of the word "net" to describe the shell just sounds to me like a poor choice of words.

 

[SeanFM21]

I don't think that's what the problem is saying. I would read that as saying that the total charge on the shell alone is +1.1uC, and then there's an additional charge of -5.3uC inside of the shell. The use of the word "net" to describe the shell just sounds to me like a poor choice of words.

I took a guess at what you were saying, so I ended up getting the problem correctly. May I ask a theoretical/problem solving question for situations like these? Is it always as simple as this problem actually was once I realized I was just misinterpreting the numbers they gave me? Is Gauss Law as simple as finding the total charge enclosed and applying other given conditions to it (depending on the shape of the surface)? In retrospect, this just seems like a very basic problem, but would I be wrong to assume they don't get much harder than this?

 

[Chopin]

Well, they can definitely get trickier than this, but the basic idea behind Gauss's Law problems is always the same: the total field (integrated across the whole surface) is proportional to the total enclosed charge. That makes the problem pretty simple, as long as you can make some deductions about how the total field relates to the field at one point.

For instance, in a problem like this, you know that the field will be spherically symmetric (since the charges are all spherically symmetric). So you know that the field will point straight in/out from the center of the sphere, and be identical in magnitude everywhere. That means that the field at any point is just equal to the total field divided by the area of the sphere, which is why you divided by 4\pi r^2.

If the problem hadn't had spherical symmetry, it would have been trickier to relate the total surface integral (which is what Gauss's Law tells you) to the strength of the field at one point. The real trick with a Gauss's Law problem is always in picking a surface whose shape will allow you to easily set up that relation between total surface integral and value at the point of interest.