A hollow conducting sphere of radius 2R is initially charged to potential V and a separate solid conducting sphere of radius R to potential V/2. The smaller sphere is placed inside the larger one without altering their respective charges. Their resulting potential difference is V/n. Find n.
Physics · Mixed Chapter
JEE Mains

Question

A hollow conducting sphere of radius 2R is initially charged to potential V and a separate solid conducting sphere of radius R to potential V/2. The smaller sphere is placed inside the larger one without altering their respective charges. Their resulting potential difference is V/n. Find n.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Initial charge of inner solid sphere: For an isolated conducting sphere, V = q/(4πε₀R). The small sphere (radius R) is at initial potential V/2, so its charge is q = 4πε₀R(V/2) = 2πε₀RV. Step 2 — Bring it concentrically inside the outer shell: The charge of each sphere remains unchanged. The potential difference between inner sphere (R) and outer shell (2R) is ΔV = [q/(4πε₀)](1/R − 1/2R). Step 3 — Substitute: ΔV = [2πε₀RV/(4πε₀)](1/2R) = V/4. So ΔV = V/n ⇒ n = 4. Answer: 4.