QJEEM23Ph072c: JEE Main 2023 Physics - Class 12 JEEM23Ph072c Solution: Three concentric spherical metallic shells X, Y and Z of radius …

Given three concentric spherical shells X, Y, and Z with radii a, b, and c respectively, and with surface charge densities ( $\sigma$ ), ( $-\sigma$ ), and ( $\sigma$ ) respectively, we know that the potential at the surface of a sphere due to a uniform surface charge is given by:

$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$

where ( $\epsilon_0$ ) is the permittivity of free space, ( Q ) is the total charge on the sphere, and ( r ) is the radius of the sphere.

However, in this case, the total charge on each sphere is given by its surface charge density ( $\sigma$ ) times its surface area ( $4\pi r^2$ ). Substituting this into the formula for ( Q ) gives:

$Q = \sigma 4\pi r^2$

So the potential at the surface of each sphere is given by:

$V = \frac{1}{4\pi\epsilon_0} \frac{\sigma 4\pi r^2}{r} = \frac{\sigma r}{\epsilon_0}$

We are given that the potential at X and Z are the same. Thus:

$V_X = V_Z$

Substituting the formula for the potential into this equation gives:

$\frac{\sigma a}{\epsilon_0} = \frac{\sigma c}{\epsilon_0}$

This simplifies to:

$a = c$

However, we also need to take into account the effect of the charge on shell Y on the potentials at X and Z. The potential at any point due to a charged shell is the same everywhere outside the shell, so we can add the potential due to shell Y at X to both sides of the equation. This gives:

$\frac{\sigma a}{\epsilon_0} - \frac{\sigma b}{\epsilon_0} + \frac{\sigma c}{\epsilon_0} = \frac{\sigma a}{\epsilon_0} + \frac{\sigma c}{\epsilon_0}$

This simplifies to:

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

Further simplification gives:

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

So:

$c = a + b$

Given that the radii of X & Y are 2 cm and 3 cm, respectively, we have:

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

Therefore, the radius of shell Z is 5 cm.