1. Given data

Three concentric metallic spherical shells:

• Shell $X$ of radius $a$
• Shell $Y$ of radius $b$
• Shell $Z$ of radius $c$

with $a<b<c$.

Surface charge densities are respectively:

• On $X$: $\sigma$
• On $Y$: $-\sigma$
• On $Z$: $\sigma$

Also given:

• $a=2\text{ cm}$
• $b=3\text{ cm}$
• Potentials of shells $X$ and $Z$ are equal.

We need to find $c$.

2. Charge on each shell

Since surface charge density is given, total charge on each shell is:

$Q_X = 4\pi a^2\sigma$ $Q_Y = -4\pi b^2\sigma$ $Q_Z = 4\pi c^2\sigma$

3. Potential at shell $X$

Potential due to a spherical shell:

• At an external point: $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$
• At any internal point/on inner region of shell: $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$ where $R$ is shell radius.

At $r=a$ (shell $X$):

• Due to shell $X$ itself: $V_{XX} = \frac{1}{4\pi\varepsilon_0}\frac{Q_X}{a}$
• Due to shell $Y$ (point inside shell $Y$): $V_{YX} = \frac{1}{4\pi\varepsilon_0}\frac{Q_Y}{b}$
• Due to shell $Z$ (point inside shell $Z$): $V_{ZX} = \frac{1}{4\pi\varepsilon_0}\frac{Q_Z}{c}$

So,

$V_X = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q_X}{a}+\frac{Q_Y}{b}+\frac{Q_Z}{c}\right)$

Substitute charges:

$V_X = \frac{1}{4\pi\varepsilon_0}\left(\frac{4\pi a^2\sigma}{a}+\frac{-4\pi b^2\sigma}{b}+\frac{4\pi c^2\sigma}{c}\right)$

$V_X = \frac{1}{4\pi\varepsilon_0}(4\pi\sigma)(a-b+c)$

4. Potential at shell $Z$

At $r=c$:

• Due to shell $X$: $V_{XZ} = \frac{1}{4\pi\varepsilon_0}\frac{Q_X}{c}$
• Due to shell $Y$: $V_{YZ} = \frac{1}{4\pi\varepsilon_0}\frac{Q_Y}{c}$
• Due to shell $Z$ itself: $V_{ZZ} = \frac{1}{4\pi\varepsilon_0}\frac{Q_Z}{c}$

Thus,

$V_Z = \frac{1}{4\pi\varepsilon_0}\frac{Q_X+Q_Y+Q_Z}{c}$

Substitute charges:

$V_Z = \frac{1}{4\pi\varepsilon_0}\frac{4\pi\sigma(a^2-b^2+c^2)}{c}$

$V_Z = \frac{1}{4\pi\varepsilon_0}(4\pi\sigma)\frac{a^2-b^2+c^2}{c}$

5. Use condition $V_X = V_Z$

Since shells $X$ and $Z$ are at same potential,

$a-b+c = \frac{a^2-b^2+c^2}{c}$

Multiply by $c$:

$c(a-b+c)=a^2-b^2+c^2$

Expand left side:

$ac-bc+c^2=a^2-b^2+c^2$

Cancel $c^2$ from both sides:

$ac-bc=a^2-b^2$

Factor both sides:

$c(a-b)=(a-b)(a+b)$

Since $a\neq b$, divide by $(a-b)$:

$c=a+b$

6. Substitute values

$c=2+3=5\text{ cm}$

7. Final answer

The radius of shell $Z$ is

$\boxed{5\text{ cm}}$

8. Comparison with stored answer

Stored correct answer: $5$

Our derived answer is also $5$, so it agrees.