A parallel plate capacitor has a capacity 80 | vedclass

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Question

(D) Initial capacitance with air,$C_{	ext{air}} = 80 	imes 10^{-6} \ F = 80 \ \mu F$. Capacitance with dielectric slab,$C_d = K 	imes C_{	ext{air}

Vedclass · Accepted Answer

$45.6 	imes 10^{-3} \ C$

Vedclass · Answer

(D) Initial capacitance with air,$C_{	ext{air}} = 80 	imes 10^{-6} \ F = 80 \ \mu F$. Capacitance with dielectric slab,$C_d = K 	imes C_{	ext{air}} = 20 	imes 80 \ \mu F = 1600 \ \mu F$. When connected to a $30 \ V$ battery,the charge stored with the dielectric is $q_d = C_d 	imes V = 1600 	imes 10^{-6} 	imes 30 = 48000 \ \mu C = 48 	imes 10^{-3} \ C$. When the dielectric is removed,the new capacitance becomes $C_{	ext{air}} = 80 \ \mu F$. The new charge stored is $q_{	ext{air}} = C_{	ext{air}} 	imes V = 80 	imes 10^{-6} 	imes 30 = 2400 \ \mu C = 2.4 	imes 10^{-3} \ C$. The charge that passes through the wire is the difference between the initial charge and the final charge: $\Delta q = q_d - q_{	ext{air}} = 48 	imes 10^{-3} \ C - 2.4 	imes 10^{-3} \ C = 45.6 	imes 10^{-3} \ C$.

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