JEE Main 2023 — Electrostatics Question with Solution

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:

$V=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}$

where ( ${ϵ}_0$ ) 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:

$V=\frac{1}{4\pi {ϵ}_0}\frac{\sigma 4\pi r^2}{r}=\frac{\sigma r}{{ϵ}_0}$

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:

$\frac{\sigma a}{{ϵ}_0}=\frac{\sigma c}{{ϵ}_0}$

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:

$\frac{\sigma a}{{ϵ}_0}-\frac{\sigma b}{{ϵ}_0}+\frac{\sigma c}{{ϵ}_0}=\frac{\sigma a}{{ϵ}_0}+\frac{\sigma c}{{ϵ}_0}$

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.