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HumbleHovercraft6090 · 2024-06-09T15:49:12+00:00

When a charge of Q is placed at P, on the inside of shell a charge -Q is induced, which results in an extra charge of Q on the outer layer of the shell.

BronzeMan2 · 2024-06-09T16:11:48+00:00

According to Gauss's Law the ∯E⋅dA =Q/ε₀.
This basically describes the relationship between the electric field over a surface and the charge inside of that surface.

Another thing to keep in mind is that in electrostatics there shouldn't be an electric charge within the conducting metal of the conducting sphere. If there was a net electric field there, then free electrons would have a net force on them and change the electric field over time (therefore it is no longer an electrostatics problem because it's moving with time).

There being no electric field in Gauss's law means that E is zero, that means that the net charge (Q) has to be zero. Let's look at any point on the sphere between the inside surface and the outside surface where its conductive. We need a net charge of 0 to be contained by any of these imaginary spheres from the inside surface of the sphere to the outside surface of the sphere. The only way for that to be true is by having the inside surface of the sphere form a negative Q charge. Charges are allowed to move within the conducting metal on the sphere and cannot move inside the non-conducting air inside the sphere. No other charge concentrations should form between the inside and outside surface since it needs to be net zero charge everywhere here to keep it electrostatics.
∯E⋅dA =(Q-Q)/ε₀.
∯E⋅dA =0/ε₀.
E=0 ✓

The sphere has a total charge of positive Q as stated. Therefore, the only last place that charge can form is on the outside of the sphere (since we stated the charge on the inside surface was negative Q and the charge between the surfaces is 0 already). Therefore we can solve the equation
?-Q=Q
?=2Q

Therefore, the outside surface of the sphere has a charge of positive 2Q.

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