1.

Capacitance of a capacitor is 16 uF. If a dielectric slab of dielectric constant 8 is inserted between the plates of capacitor.Then the capacitance of the capacitor will beA) 128uFB) 16uFC) None of theseD) 2uF​

Answer»

B is CORRECT

Please mark me as brainliest

EXPLANATION

Given, 16μF V = Q/C = Q/16μF When DIELECTRIC is inserted V' = Q/C = Q/K (16μF) = V/8 Q/K (16μF) = Q/8 × 16μF K = 8 OPTION B is correct.



Discussion

No Comment Found

Related InterviewSolutions