Induced charge on conductors

[RodolfoM]

Homework Statement

In the following figure, there is a massive sphere A, concentrical to the spherical shells B and C. The spherical shell B has a positive net charge $q_B$, while C is neutral and the sphere A is connected to the ground through a thin wire, with neglectable capacitance, which doesn't touch the shells B and C. You should find:
a) The induced charges on the surfaces of the sphere A and the shells B and C.
b) The new charge of the sphere A after the shell C becomes grounded.

Relevant Equations

Gauss's law:
$\oint E dA=Q_{enc}/\varepsilon$
Poisson's Equation.

My first attempt at solving this was to calculate the induced charge in A by making $V=0$:
$\frac{q_{A}}{4πεR_{1}}+\frac{q_{B}}{4πεR_{3}}=0$
$q_{A}=-q_{B}\frac{R_{1}}{R_{3}}$

But that's not the answer. Any help is welcome!

 

[Charles Link]

I first thought you might have it correct, but then I see where your method neglects charge that gets induced on the inner surface and outer surface of the B shell. Once you take account of that and compute $V=-\int E \cdot ds$ for the stretch from $R_2$ to $R_1$ you will get a different answer.

 

[RodolfoM]

I first thought you might have it correct, but then I see where your method neglects charge that gets induced on the inner surface and outer surface of the B shell. Once you take account of that and compute $V=-\int E \cdot ds$ for the stretch from $R_2$ to $R_1$ you will get a different answer.

Yeah, I've been thinking about this problem. What do you think about the following solution?

• There will be a redistribution of the charge on sphere B, let's name $q_{B_{3}}$ the outer charge (radius $R_{3}$) and $q_{B_{2}}$ the inner charge (radius $R_{2}$).
• The inside sphere A will have an induced charge $q_{A}$.
• There should be no eletric field inside the sphere shell B, therefore we have $q_{B_{2}}=-q_{A}$, because a Gaussian surface drawn inside the shell should give us $q_{enc}=0$, according to Gauss's Law.
• The net charge of shell B will be the same: $q_{B_{2}}+q_{B_{3}}=q_{B}$.
• The potencial at the surface of sphere A is zero, therefore $\frac{Kq_{A}}{R_{1}}+\frac{Kq_{B_{2}}}{R_{2}}+\frac{Kq_{B_{3}}}{R_{3}}=0$

That gives us the following system:
$\begin{cases}\frac{q_{A}}{R_{1}}+\frac{q_{B_{2}}}{R_{2}}+\frac{q_{B_{3}}}{R_{3}}=0\\q_{B_{2}}=-q_{A}\\q_{B_{2}}+q_{B_{3}}=q_{B} \end{cases}$

Solving for $q_{A}$, we have
$q_{A}=\frac{-q_{B}}{1+\frac{R_{3}(R_{2}-R_{1})}{R_{1}R_{2}}}$

In the original exercise, the values are $q_{B}=4.0\mu C, R_{1}=10 cm, R_{2}=20 cm$ and $R_{3}=40 cm$, that would result in $q_{A}=-4/3 \mu C$. Do you think this reasoning makes sense?

 

[haruspex]

• The potencial at the surface of sphere A is zero, therefore $\frac{Kq_{A}}{R_{1}}+\frac{Kq_{B_{2}}}{R_{2}}+\frac{Kq_{B_{3}}}{R_{3}}=0$

That neglects the potential due to the outermost shell.

 

[Charles Link]

The potencial at the surface of sphere A is zero, therefore KqAR1+KqB2R2+KqB3R3=0

I think this expression is correct. I worked it by computing potentials from the electric fields in the various regions, and unless I missed something, I got an identical answer. I looked over most of your solution, and it looks right to me. I haven't yet gone over it real meticulously. Edit: My statements here are incorrect. See posts 6 and 7.

 

[haruspex]

I think this expression is correct. I worked it by computing potentials from the electric fields in the various regions, and unless I missed something, I got an identical answer. I looked over most of your solution, and it looks right to me. I haven't yet gone over it real meticulously.

You don't think the outermost shell is relevant?

 

[Charles Link]

You don't think the outermost shell is relevant?

Thank you @haruspex . I think you may be right. I ignored it in my solution, and my result may be in error. The computation of the surface charges on the outer shell is straightforward, and yes, they will affect the result.

 

[alan123hk]

In the original exercise, the values are qB=4.0μC,R1=10cm,R2=20cmqB=4.0μC,R1=10cm,R2=20cmq_{B}=4.0\mu C, R_{1}=10 cm, R_{2}=20 cm and R3=40cmR3=40cmR_{3}=40 cm, that would result in qA=−4/3μCqA=−4/3μCq_{A}=-4/3 \mu C. Do you think this reasoning makes sense?

I think that if the outermost shell is not grounded, the answer should depend on R1, R2, R3, R4, and R5; on the other hand, if the outermost shell is grounded, the answer should only depend on R1, R2, R3, and R4.

The charge on shell B will be redistributed on its inner and outer surfaces, and their ratio should be related to R1, R2, R3, R4 ...

Imagine that if the outermost shell is not grounded and R4 is very close to R3, then almost all the charge should be distributed on the outer surface of shell B, because the capacitance of R3-R4 is much larger than that of R2-R1.

 

[haruspex]

I think that if the outermost shell is not grounded, the answer should depend on R1, R2, R3, R4, and R5; on the other hand, if the outermost shell is grounded, the answer should only depend on R1, R2, R3, and R4.

The charge on shell B will be redistributed on its inner and outer surfaces, and their ratio should be related to R1, R2, R3, R4 ...

Imagine that if the outermost shell is not grounded and R4 is very close to R3, then almost all the charge should be distributed on the outer surface of shell B, because the capacitance of R3-R4 is much larger than that of R2-R1.

Perhaps, but is all that borne out by equations?

 

[Charles Link]

Perhaps, but how about some equations?

Hopefully the OP returns and shows us what he came up with.

 

[haruspex]

Hopefully the OP returns and shows us what he came up with.

Thanks for the hint that I was not replying to the OP. Post edited.

 

[rude man]

I would start with $V_C$ and work my way towards $V_A$, integrarung E fields along the way, yielding $q_A$.

 

[alan123hk]

Perhaps, but is all that borne out by equations?

I think from an engineering perspective, the easiest way to solve this problem is to use the concept of capacitance.

By applying Gauss's law to a charged conductive sphere, the capacitance between the spheres can be obtained effortlessly.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html

Assuming that the outermost shell C is grounded, since the sphere A is also grounded, the entire structure can be regarded as two capacitors C_cb and C_ba connected in parallel.

C_cb also represents the capacitance between outermost shell C and shell B
C_ba also represents the capacitance between shell B and sphere A

Thus the equations are : -

V_b C_cb = Q_{b(outer surface)}
V_b C_ba = Q_{b(inner surface)}
Q_b = Q_{b(outer surface)} + Q_{b(inner surface)}

C_cb, C_ba and Q_b are known
V_b, Q_{b(outer surface)} and Q_{b(inner surface)} are unknown

For the case where the outermost shell C is not grounded, we can solve the problem in a similar manner.

 

[rude man]

V_b C_cb = Q_{b(outer surface)}

I think $(V_B - V_C)C_{cb} = Q_{b~ outer~ surface}$

 

[alan123hk]

I think $(V_B - V_C)C_{cb} = Q_{b~ outer~ surface}$

Yes, the general form should be (V_b−V_c)C_cb=Q_{b(outer surface)}

Questions in the original post
a) The induced charges on the surfaces of the sphere A and the shells B and C.
b) The new charge of the sphere A after the shell C becomes grounded.

I just assume V_c=0 to fulfill the condition of question b), also Q_a should be equal to -Q_{b(inner surface)}.

By applying the general form (V_b−V_c)C_cb=Q_{b(outer surface)}, we can solve a similar system of equations for any voltage of V_c.