A parallel plate capacitor with air between the plates has a capacitance of 9 pF . The separation between the plates is d . The space between the plates is now filled with two dielectrics constant K 1 = 3 and thickness d 3 while the other one has dielectric constant K 2 = 6 and thickness 2 d 3 . Capacitance of the capacitor is now

Question

A parallel plate capacitor with air between the plates has a capacitance of 9 pF . The separation between the plates is d . The space between the plates is now filled with two dielectrics constant K 1 = 3 and thickness d 3 while the other one has dielectric constant K 2 = 6 and thickness 2 d 3 . Capacitance of the capacitor is now, 1 . 8 pF, 45 pF, 40 . 5 pF, 20 . 25 pF

MARKS App · Accepted Answer

The two capacitors formed are in parallel, hence capacitance of the combination C = C 1 C 2 C 1 + C 2 . . . ( i ) where, C 1 = K 1 ε 0 A d 3 ∵ C = ε 0 A d . . . ( ii ) C 2 = K 2 ε 0 A 2 d 3 . . . ( iii ) C eq = 3 K 1 ε 0 A d × K 2 ε 0 A · 3 2 d 3 K 1 ε 0 A d + 3 K 2 ε 0 A 2 d = 9 K 1 K 2 ε 0 2 A 2 2 d 2 × d ε 0 A × 18 It is given ε 0 A d = 9 pF Using given values C eq = 40 . 5 pF

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