# Homework Help: Electrostatic equilibrium (Sphere with cavity)

1. Nov 7, 2012

### Fabio010

1. The problem statement, all variables and given/known data

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?

3. The attempt at a solution

a)
Qint = -$\sum$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 = -$\sum$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 = -$\sum$qi

2.0 * 10^-9 + x = -$\sum$qi

2.0 * 10^-9 - 1.2E-9 = -$\sum$qi

-$\sum$qi = 8 E -10C

Qint = 8 E -10 C = Qext

Last edited by a moderator: Nov 7, 2012
2. Nov 7, 2012

### haruspex

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

3. Nov 7, 2012

### Fabio010

sorry you right.

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

$\sum$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)

4. Nov 8, 2012

### haruspex

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?