What is the charge on the insulating sphere?

[Maltesers123]

For the configuration shown in the figure below, suppose a = 5.00 cm, b = 20.0 cm, and c = 25.0 cm. Furthermore, suppose the electric field at a point 10.5 cm from the center is measured to be 3.25 × 10^3 N/C radially inward and the electric field at a point 50.0 cm from the center is of magnitude 176 N/C and points radially outward. From this information, find the following. (Include the sign of the charge in your answer.)

(a) the charge on the insulating sphere

i have calculated as follow:
E( the surface area of sphere) = q(inside)/ permittivity
3.25χ10^3 (4χ3.16χ0.105^2)= q(inside)/ (8.85χ10^-12)
q(inside) = -0.000000004 C

this is the correct answer but i don't know why my equation is valid.

First, there is charge in the insulator and the inner wall of the conductor, right?

and isn't the electric field measured at 10.5 cm from the center of the insulator the sum of the effect from both the charge from the insulator and the inner wall of the conductor?

if it is true, then why i can use the gauss law to find the total charge inside the insulator like this?

 

[Doc Al]

First, there is charge in the insulator and the inner wall of the conductor, right?

Sure. And maybe the outer wall of the conductor as well.

and isn't the electric field measured at 10.5 cm from the center of the insulator the sum of the effect from both the charge from the insulator and the inner wall of the conductor?

The field at any point is due to all the charges.

if it is true, then why i can use the gauss law to find the total charge inside the insulator like this?

Symmetry allows you to not worry about charge at any point where r > 10.5 cm. The effect of those charges cancels out. So all you need to care about are the charges within the insulator.

 

[Maltesers123]

Symmetry allows you to not worry about charge at any point where r > 10.5 cm. The effect of those charges cancels out. So all you need to care about are the charges within the insulator.

thank you for your answering, but i don't understand why the symmetry allows tot cancels out the effect of those charges.

at 10 .5 cm , the distance from one side of the charge of inner surface of the conductor is different from the opposite side. electric varies with the distance, doesn't the charges still contribute to the electric field at 10.5cm?

 

[Doc Al]

thank you for your answering, but i don't understand why the symmetry allows tot cancels out the effect of those charges.

at 10 .5 cm , the distance from one side of the charge of inner surface of the conductor is different from the opposite side. electric varies with the distance, doesn't the charges still contribute to the electric field at 10.5cm?

It turns out, and this is encapsulated within Gauss's law, that the contribution of the charges where r > 10.5 cm exactly cancels. Newton proved this in one of his Shell Theorems.

Say you have a uniform spherical shell of charge. Outside the shell, the field is exactly that of a point charge located at the center of the shell with the same total charge. Within the shell, the field everywhere is zero. (Same thing is true for gravity or any inverse square law force.)

 

[jilia]

why do you have negative sign for the q(inside)? can you explain?

 

[Doc Al]

why do you have negative sign for the q(inside)? can you explain?

Do you mean the q on the insulating sphere? Note that the field at 10.5 cm from the center is radially inward.