JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Electrostatics

Before the extra charge is added the configuration behaves like two concentric spheres, so the potential difference between the inner sphere and the outer surface of the shell is $V_{\text{in}}-V_{\text{out}}=kQ\left(\frac{1}{r_1}-\frac{1}{r_2}\right)=V$, a quantity fixed by the inner charge $Q$ and the two radii alone. When the shell is handed a charge $-4Q$, that charge raises or lowers the potential of every point at and inside the shell by the same constant amount $-\frac{4kQ}{r_2}$. Both surfaces shift by exactly this common constant, so the difference between them is untouched and the potential difference stays $V$. The lesson worth carrying away is that charge placed on the enclosing shell shifts potentials but never alters the inner-to-outer difference, which the inner charge sets by itself.