A parallel-plate capacitor carries charge 5×10⁻⁶ C. A dielectric slab is inserted between the plates and nearly fills the space. The induced charge on one face of the slab is 4×10⁻⁶ C. Find the dielectric constant K.
Physics · Capacitor · capacitor with dielectric
JEE Mains 2025 7 April Evening Shift

Question

A parallel-plate capacitor carries charge 5×10⁻⁶ C. A dielectric slab is inserted between the plates and nearly fills the space. The induced charge on one face of the slab is 4×10⁻⁶ C. Find the dielectric constant K.

Physics · Capacitor · capacitor with dielectric

JEE Mains 2025 7 April Evening Shift

G1 Physics · Accepted Answer

For a dielectric filling a capacitor, the induced bound charge magnitude is Q i n d Q_ind Q i ​ n d = Q(1-1/K), where Q is the free charge on the capacitor plates. Given, 4×10⁻⁶ = 5×10⁻⁶(1-1/K). Cancel 10⁻⁶: 4 = 5(1-1/K). Thus, 4/5 = 1-1/K ⇒ 1/K = 1/5 ⇒ K = 5. Answer: 5.