Gauss' law -- Conductor with a cavity

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Physicslearner500039
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Homework Statement
An isolated conductor has net charge +10 X 10 -6 C and a cavity with a point charge q = +3.0 X 10-6 C. What is the charge on (a) the cavity wall and (b) the outer surface?
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I really don't understand the theory of the above kind of questions. But from the little theory i understand the Electric field is 0 inside the conductor and all the charge goes to the surface and distributes equally.

a. Since the E=0 inside the conductor the point charge distributes outside the cavity to make the net charge as 0. Hence the cavity develops a charge of -3uC on its surface. I am not 100% sure how it works and the negative sign. Please advise.

b. If i apply the Gauss Law with a surface around the conductor the net charge enclosed is 10uC + 3uC = 13uC distributes on the surface of the conductor.
 
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Physicslearner500039 said:
  1. Electric field is 0 inside the conductor ##\qquad## correct
  2. and all the charge goes to the surface ##\qquad## correct (surfaces)
  3. and distributes equally ##\qquad## ##\qquad## ##\qquad\quad## not correct
ad 2: there can be more surfaces, like in this exercise
ad 3: equally distributed happens in specific cases (with some symmetry). The basic rule is that the electric field lines are perpendicular to the surface(s)
(if not, there would be a component along the surface, and the charge would re-distribute until that component is zero everywhere).

In your exercise it is not given that the cavity is in the centre of the sphere, nor that the charge in in the centre of the cavity. But you can, as @hutchphd indicates, make good use of Gauss' theorem.
 
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I have drawn a diagram to understand, the q1 = 3uC. If we consider the gaussian surface the Flux is zero
∅ = 0; ∈0*∅ = q_enclosed;
q_enclosed = 0;
Let the charge on the cavity is q_cavity;
q_cavity + q1 = 0;
q_cavity = -3uC
This i understood. The confusion is for the net charge let us say q_net.
The same principle i apply ∅ = 0;
The charge on the surface is q_surface;
q_surface + qnet +q_cavity+q1 = 0;
q_surface = -qnet; = -10uC; The answer is wrong and i am completely wrong, where i am making the mistake. Please advise.
 
Physicslearner500039 said:
q_cavity = -3uC
Correct.
The conductor charge is +10 ##\mu##C.
What is left of the conductor charge is sitting on the outside of the conductor. How much is that if you subtract the -3 ##\mu##C from the original +10 ##\mu##C ?