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A point charge $q$ is at the center of an uncharged conducting spherical shell of finite thickness, with inner and outer radii $a$ and $b$ respectively. Find the work done on the system when $q$ is removed from its original position to a very large distance from the conducting shell, through a small hole in it.

My attempt:

Idea was to calculate the energy of the initial configuration and negative of that would be the work required. Trying to use $ W = \frac {\epsilon_0}{2} \int E^2 d\tau $ didn't work as the integral goes to infinity at r=0. So I used $ W =\frac 12 \int \sigma V ds$ where $\sigma$ is the induced charge on the two surface at radii a and b.

$$ W= \frac {q}{8\pi \epsilon_0b} (\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da ) $$
which come out to be zero as each integral is equal to $q$ in magnitude and opposite in sign. I think this is wrong. How do I solve this problem?

EDIT: I think I forgot the terms due to the potential of induced charges at the center. $$ W= \frac {q}{4\pi \epsilon_0b} (\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da ) + \frac {q^2}{4\pi \epsilon_0b} - \frac {q^2}{4\pi \epsilon_0a} $$ This comes out as $$ W= \frac {q^2}{4\pi \epsilon_0}(\frac1b - \frac1a) $$ External work done is negative of this so that a positive work is done in taking the charge to infinity. Is this correct?

A point charge $q$ is at the center of an uncharged conducting spherical shell of finite thickness, with inner and outer radii $a$ and $b$ respectively. Find the work done on the system when $q$ is removed from its original position to a very large distance from the conducting shell, through a small hole in it.

My attempt:

Idea was to calculate the energy of the initial configuration and negative of that would be the work required. Trying to use $ W = \frac {\epsilon_0}{2} \int E^2 d\tau $ didn't work as the integral goes to infinity at r=0. So I used $ W =\frac 12 \int \sigma V ds$ where $\sigma$ is the induced charge on the two surface at radii a and b.

$$ W= \frac {q}{8\pi \epsilon_0b} \left(\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da \right) $$
which come out to be zero as each integral is equal to $q$ in magnitude and opposite in sign. I think this is wrong. How do I solve this problem?

EDIT: I think I forgot the terms due to the potential of induced charges at the center. $$ W= \frac {q}{4\pi \epsilon_0b} \left(\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da \right) + \frac {q^2}{4\pi \epsilon_0b} - \frac {q^2}{4\pi \epsilon_0a} $$ This comes out as $$ W= \frac {q^2}{4\pi \epsilon_0} \left(\frac1b - \frac1a \right) $$ External work done is negative of this so that a positive work is done in taking the charge to infinity. Is this correct?

A point charge $q$ is at the center of an uncharged conducting spherical shell of finite thickness, with inner and outer radii $a$ and $b$ respectively. Find the work done on the system when $q$ is removed from its original position to a very large distance from the conducting shell, through a small hole in it.

My attempt:

Idea was to calculate the energy of the initial configuration and that would be the work required. Trying to use $ W = \frac {\epsilon_0}{2} \int E^2 d\tau $ didn't work as the integral goes to infinity at r=0. So I used $ W =\frac 12 \int \sigma V ds$ where $\sigma$ is the induced charge on the two surface at radii a and b.

$$ W= \frac {q}{8\pi \epsilon_0b} (\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da ) $$
which come out to be zero as each integral is equal to $q$ in magnitude and opposite in sign. I think this is wrong. How do I solve this problem?

A point charge $q$ is at the center of an uncharged conducting spherical shell of finite thickness, with inner and outer radii $a$ and $b$ respectively. Find the work done on the system when $q$ is removed from its original position to a very large distance from the conducting shell, through a small hole in it.

My attempt:

Idea was to calculate the energy of the initial configuration and negative of that would be the work required. Trying to use $ W = \frac {\epsilon_0}{2} \int E^2 d\tau $ didn't work as the integral goes to infinity at r=0. So I used $ W =\frac 12 \int \sigma V ds$ where $\sigma$ is the induced charge on the two surface at radii a and b.

$$ W= \frac {q}{8\pi \epsilon_0b} (\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da ) $$
which come out to be zero as each integral is equal to $q$ in magnitude and opposite in sign. I think this is wrong. How do I solve this problem?

EDIT: I think I forgot the terms due to the potential of induced charges at the center. $$ W= \frac {q}{4\pi \epsilon_0b} (\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da ) + \frac {q^2}{4\pi \epsilon_0b} - \frac {q^2}{4\pi \epsilon_0a} $$ This comes out as $$ W= \frac {q^2}{4\pi \epsilon_0}(\frac1b - \frac1a) $$ External work done is negative of this so that a positive work is done in taking the charge to infinity. Is this correct?

A point charge $q$ is at the center of an uncharged conducting spherical shell of finite thickness, with inner and outer radii a and b respectively. Find the work done on the system when $q$ is removed from its original position to a very large distance from the conducting shell, through a small hole in it.

My attempt:

Idea was to calculate the energy of the initial configuration and that would be the work required. Trying to use $ W = \frac {\epsilon_0}{2} \int E^2 d\tau $ didn't work as the integral goes to infinity at r=0. So I used $ W =\frac 12 \int \sigma V ds$ where $\sigma$ is the induced charge on the two surface at radii a and b.

$$ W= \frac {q}{8\pi \epsilon_0b} (\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da ) $$
which come out to be zero as each integral is equal to $q$ in magnitude and opposite in sign. I think this is wrong. How do I solve this problem?

A point charge $q$ is at the center of an uncharged conducting spherical shell of finite thickness, with inner and outer radii $a$ and $b$ respectively. Find the work done on the system when $q$ is removed from its original position to a very large distance from the conducting shell, through a small hole in it.

My attempt:

Idea was to calculate the energy of the initial configuration and that would be the work required. Trying to use $ W = \frac {\epsilon_0}{2} \int E^2 d\tau $ didn't work as the integral goes to infinity at r=0. So I used $ W =\frac 12 \int \sigma V ds$ where $\sigma$ is the induced charge on the two surface at radii a and b.

$$ W= \frac {q}{8\pi \epsilon_0b} (\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da ) $$
which come out to be zero as each integral is equal to $q$ in magnitude and opposite in sign. I think this is wrong. How do I solve this problem?