Some problems in electic field

[wowowo2006]

hey i got some problems in doing this question
http://i4.photobucket.com/albums/y117/tkm20022001/AL%20PHY/phy03mc_32_33.jpg"
the official ans is A
i don't know why the 2nd statement is wrong. i think it's right as it;s at equilibrium state if not the charge will accelerate due to e field presents and may affect the charge distribution which is not at a equilibrium
moreover i don't know why the 1st statement is true, the - charges should be denser at the inner surface near the + charge, why there is still evenly distrubuted + charges on outer surface denser even when the - charges are denser at some place.
i am just studying some basic physics
and my mates explain this to me with shielding effect and gauss surface. i don't know about this . Could someone help me solve this question and give me some simple explanation about shielding effect and gauss surface and how they can help me get the answer.

 

[Doc Al]

i don't know why the 2nd statement is wrong. i think it's right as it;s at equilibrium state if not the charge will accelerate due to e field presents and may affect the charge distribution which is not at a equilibrium

Think of the charge within the cavity as being fixed in place by other forces. When they speak of electrostatic equilibrium, they mean all the charges within and on the conducting shell.

moreover i don't know why the 1st statement is true, the - charges should be denser at the inner surface near the + charge, why there is still evenly distrubuted + charges on outer surface denser even when the - charges are denser at some place.

The charges on the inner surface arrange themselves to cancel any field from the charge within the cavity--the field within the conducing shell must be zero. That effectively shields the outer surface from any effect of the charge within. Since it's a spherical surface, the induced charge on the outer surface distributes itself uniformly.

 

[wowowo2006]

Think of the charge within the cavity as being fixed in place by other forces. When they speak of electrostatic equilibrium, they mean all the charges within and on the conducting shell.

The charges on the inner surface arrange themselves to cancel any field from the charge within the cavity--the field within the conducing shell must be zero. That effectively shields the outer surface from any effect of the charge within. Since it's a spherical surface, the induced charge on the outer surface distributes itself uniformly.

sorry i don't know what does 'The charges on the inner surface arrange themselves to cancel any field from the charge within the cavity' mean. does it mean that the e field within the cavity and the metal shell , both e field is zero or only with the shell. and why does the 'self arrangement of + charges' is uniform, is it just because the ball is a perfect sphere?. i don't know exactly why the unevenly distributed - charges can in a equilibrium state when the + charges besides them is evenly distributed. Could you give me more explanations please

 

[Doc Al]

sorry i don't know what does 'The charges on the inner surface arrange themselves to cancel any field from the charge within the cavity' mean. does it mean that the e field within the cavity and the metal shell , both e field is zero or only with the shell.

Sorry about that. I meant that the charges on the inner surface of the shell arrange themselves to ensure that the field is zero throughout the material of the shell (and all points beyond), not within the cavity itself.

and why does the 'self arrangement of + charges' is uniform, is it just because the ball is a perfect sphere?. i don't know exactly why the unevenly distributed - charges can in a equilibrium state when the + charges besides them is evenly distributed.

The combined field of the charge within the cavity and the charge on the inner surface of the shell is zero for all points beyond the inner surface. Thus as far as the outer surface goes, the charges on the inside exert no effect.

 

[PJC01]

I understand your confusion with this question. Let me help clarify some of the concepts for you.

Firstly, the 1st statement is true because when an external electric field is applied to a conductor, the charges inside the conductor will redistribute themselves in such a way that the electric field inside the conductor is equal to zero. This is known as the electrostatic shielding effect. In this case, the - charges will indeed be denser near the inner surface near the + charge, but this redistribution also causes the + charges to be evenly distributed on the outer surface, resulting in a net zero electric field inside the conductor.

Now, for the 2nd statement, you are correct in saying that the charge will accelerate due to the presence of an electric field. However, in this scenario, the charge is at equilibrium because it is held in place by the conducting material. This means that the charge is not accelerating and therefore, there is no electric field present inside the conductor. This is why the 2nd statement is incorrect.

The Gauss surface mentioned in your question is a hypothetical surface used in electromagnetism to calculate the electric field at a specific point. It is a closed surface that is chosen such that the electric field at every point on the surface is constant and perpendicular to the surface. This allows for easier calculation of the electric field using Gauss's Law.

I hope this helps to clarify the concepts of the electrostatic shielding effect and Gauss surface for you. Remember, in physics, it is important to understand the underlying principles and concepts rather than just memorizing formulas. Keep studying and don't hesitate to ask for help when needed.