- #1
- 2,567
- 4
Problem: Two spherical cavities, of radii a and b, are hollowed out from the interior of a (neutral) conducting sphere of radius R. At the center of each cavity a point charge is placed -- call these carges [itex]q_a[/itex] and [itex]q_b[/itex].
(a) Find the surface charges [itex]\sigma _a[/itex], [itex]\sigma _b[/itex], and [itex]\sigma _R[/itex].
(b) What is the field outside the conductor?
(c) What is the filed within each cavity?
(d) What is the force on [itex]q_a[/itex] and [itex]q_b[/itex]?
(e) Which of these answers would change if a third charge, [itex]q_c[/itex], were brought near the conductor?
-------------------------------------------------------------------
(a) I think [itex]q_a[/itex] would induce a charge of [itex]q_a[/itex] on the surface of the sphere, but this would include the surface of cavity b. At the same time, [itex]q_b[/itex] would induce a charge of [itex]-q_b[/itex] on the surface of its own cavity, so:
[tex]\sigma _b = \frac{q_a - q_b}{4\pi b^2}[/tex]
Similarly
[tex]\sigma _a = \frac{q_b - q_a}{4\pi a^2}[/tex]
and
[tex]\sigma _R = \frac{q_a + q_b}{4\pi R^2}[/tex]
(b) [tex]\mathbf{E}(\mathbf{r}) = \frac{q_a + q_b}{4\pi \epsilon _0 r^3}\mathbf{r}[/tex]
(c) We can treat cavity A as a spherical shell of radius a centered at the origin. The potential inside a spherical shell with uniform surface charge distribution is constant, so the electric field is zero due to the surface charge. Therefore, the only charge that matters is the point charge in the center. So the field inside is:
[tex]\mathbf{E}(\mathbf{r}) = \frac{q_a}{4\pi \epsilon _0 r^3}\mathbf{r}[/tex]
Except at the very center, since the field of the charge doesn't apply to itself, so the field there would be 0.
(d) All effects that [itex]q_a[/itex] would have on [itex]q_b[/itex] are transmitted via the surface of cavity b, just as the effect on any particle outside the big sphere from the charges inside is "communicated" via the surface of the sphere. The position of the internal charges and the shapes of the cavities don't effect points outside the sphere, all that matters is the surface charge that ends up on the sphere. Similarly, the only thing that will affect [itex]q_b[/itex] is the surface it faces (and nothing beyond it), and since the surface around it is a sphere with a uniform charge distribution, and since [itex]q_b[/itex] is at the center of this sphere, the forces on it should cancel, and the net force on it should be 0; likewise for [itex]q_a[/itex].
Also, since the field at the center is zero, the force is [itex]0 \times q_a = 0[/itex].
(e) All except the last one.
-------------------------------------------------------------------
Have I done the above correctly? Our book talks a bit about how things work with one cavity, so the above answers are based on how I think those facts would extend to the case where there are 2 cavities, but I'm not entirely sure of the above, so please let me know, and if I've done anything wrong, let me know what's wrong about it, thanks.
(a) Find the surface charges [itex]\sigma _a[/itex], [itex]\sigma _b[/itex], and [itex]\sigma _R[/itex].
(b) What is the field outside the conductor?
(c) What is the filed within each cavity?
(d) What is the force on [itex]q_a[/itex] and [itex]q_b[/itex]?
(e) Which of these answers would change if a third charge, [itex]q_c[/itex], were brought near the conductor?
-------------------------------------------------------------------
(a) I think [itex]q_a[/itex] would induce a charge of [itex]q_a[/itex] on the surface of the sphere, but this would include the surface of cavity b. At the same time, [itex]q_b[/itex] would induce a charge of [itex]-q_b[/itex] on the surface of its own cavity, so:
[tex]\sigma _b = \frac{q_a - q_b}{4\pi b^2}[/tex]
Similarly
[tex]\sigma _a = \frac{q_b - q_a}{4\pi a^2}[/tex]
and
[tex]\sigma _R = \frac{q_a + q_b}{4\pi R^2}[/tex]
(b) [tex]\mathbf{E}(\mathbf{r}) = \frac{q_a + q_b}{4\pi \epsilon _0 r^3}\mathbf{r}[/tex]
(c) We can treat cavity A as a spherical shell of radius a centered at the origin. The potential inside a spherical shell with uniform surface charge distribution is constant, so the electric field is zero due to the surface charge. Therefore, the only charge that matters is the point charge in the center. So the field inside is:
[tex]\mathbf{E}(\mathbf{r}) = \frac{q_a}{4\pi \epsilon _0 r^3}\mathbf{r}[/tex]
Except at the very center, since the field of the charge doesn't apply to itself, so the field there would be 0.
(d) All effects that [itex]q_a[/itex] would have on [itex]q_b[/itex] are transmitted via the surface of cavity b, just as the effect on any particle outside the big sphere from the charges inside is "communicated" via the surface of the sphere. The position of the internal charges and the shapes of the cavities don't effect points outside the sphere, all that matters is the surface charge that ends up on the sphere. Similarly, the only thing that will affect [itex]q_b[/itex] is the surface it faces (and nothing beyond it), and since the surface around it is a sphere with a uniform charge distribution, and since [itex]q_b[/itex] is at the center of this sphere, the forces on it should cancel, and the net force on it should be 0; likewise for [itex]q_a[/itex].
Also, since the field at the center is zero, the force is [itex]0 \times q_a = 0[/itex].
(e) All except the last one.
-------------------------------------------------------------------
Have I done the above correctly? Our book talks a bit about how things work with one cavity, so the above answers are based on how I think those facts would extend to the case where there are 2 cavities, but I'm not entirely sure of the above, so please let me know, and if I've done anything wrong, let me know what's wrong about it, thanks.