Electrostatic equilibrium (Sphere with cavity)

[Fabio010]

Homework Statement

Consider a sphere conductor with a inner concentric sphere cavity. The radius of the inside and outside of sphere are 10 cm and 20 cm. The electric field in a point P, along the outer surface has a intensity of 450 N/C and it is directed out of the surface. When a particle of unknown charge is introduced into the center cavity, the electric field continues to point outward from the surface but its intensity is now 180N/C.

a) Which was the total charge of the sphere before being introduced charge Q?

b) Which is the value of charge Q?

c) After the introduction of the charge Q which is the value of the charge on the
inner surfaces and outside conductor?

The Attempt at a Solution

a)
Qint = -[itex]\sum[/itex]qi

qi = charge inside cavity

Qint = charge between 10cm to 20cm

Qext = Outside charge

Q = Qint = Qext

The Electric field outside sphere is E = k*q/r^2


so

450 = k*q/(0.2)^2

q = 2.0 * 10^-9 C

then Qint=Qext 2.0 * 10^-9 C

b)

Qint = -[itex]\sum[/itex]qi


E = k*q/r^2

180 = k* (2.0 * 10^-9 + x)/(0.2)^2

x = -1.2E-9 C

The introduced charge in cavity is 1.2E-9 C


c)

Qint = -[itex]\sum[/itex]qi

2.0 * 10^-9 + x = -[itex]\sum[/itex]qi

2.0 * 10^-9 - 1.2E-9 = -[itex]\sum[/itex]qi

-[itex]\sum[/itex]qi = 8 E -10C

Qint = 8 E -10 C = Qext

 

[haruspex]

Qint = charge between 10cm to 20cm
Qext = Outside charge
Q = Qint = Qext

Qint is the charge on the inner surface, no? It won't equal Qext. Did you mean Q = Qint+Qext?

c)
Qint = -[itex]\sum[/itex]qi
2.0 * 10^-9 + x = -[itex]\sum[/itex]qi
2.0 * 10^-9 - 1.2E-9 = -[itex]\sum[/itex]qi
-[itex]\sum[/itex]qi = 8 E -10C
Qint = 8 E -10 C = Qext

I don't understand your calculation here. Isn't Qint just equal and opposite to the introduced charge? And Qext is whatever is left over from the original charge?

 

[Fabio010]

sorry you right.


The electric potential outside sphere is ke*Q/r^2

[itex]\sum[/itex]qi are the sum of the charges inside the sphere.

The outside charge is Qext

450 = k*q/(0.2)^2

Qext = 2.0 * 10^-9 C


I just said that the Qint in always the same. The electric field is the only that is diffrent because it varies with the radius.

so Qint (inner surface (10-20 cm) is equal to Qext (20 to infinity)

right?


When the outside electric field(along the surface = 0.2 cm) is 180, then:

180(4*pi*(0.2)^2) = Q/εo

so Q is now 28.8*pi*εo C


The introduced charge is positive


is the difference ( 2.0 * 10^-9 C - 28.8*pi*εo C)

 

[haruspex]

The electric potential outside sphere is ke*Q/r^2

[itex]\sum[/itex]qi are the sum of the charges inside the sphere.
The outside charge is Qext
450 = k*q/(0.2)^2
Qext = 2.0 * 10^-9 C

I just said that the Qint in always the same. The electric field is the only that is diffrent because it varies with the radius.

so Qint (inner surface (10-20 cm) is equal to Qext (20 to infinity)

I'm finding your use of variables confusing still.
Before the introduced charge, there will be a charge of 2.0 * 10^-9 C on the external surface of the sphere (call this Qext, but this is right at 20cm, not 20 to infinity). Let's write Qint = 0 to represent that there is no charge on the internal surface (i.e. at 10cm, not 10-20 cm) at this time.
After the charge is introduced, there will be different charge Qext' on the external surface and a nonzero charge, Qint' on the internal surface.
What will Qint' be, and having determined that what will Qext' be?

 

[Nickie]

After the introduction of the charge Q, the charge on the inner and outer surfaces of the conductor will be equal to 8 E -10 C. This is because the total charge of the sphere remains the same, and the introduced charge in the cavity will induce an equal and opposite charge on the inner and outer surfaces of the conductor in order to maintain electrostatic equilibrium. This phenomenon is known as charge redistribution.