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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 |
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Answer» 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. |
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