A parallel plate capacitor has a capacity 80 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Initial capacitance with air is $C = 80 	imes 10^{-6} \,F$.When the dielectric slab of constant $k = 20$ is inserted, the new capacitance becomes

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Initial capacitance with air is $C = 80 	imes 10^{-6} \,F$.When the dielectric slab of constant $k = 20$ is inserted, the new capacitance becomes $C' = kC = 20 	imes 80 	imes 10^{-6} = 1600 	imes 10^{-6} \,F$.The capacitor is connected to a battery of $V = 30 \,V$, so the charge on the capacitor is $q' = C'V = (1600 	imes 10^{-6}) 	imes 30 = 48000 	imes 10^{-6} \,C = 48 	imes 10^{-3} \,C$.When the dielectric slab is removed, the capacitance returns to $C = 80 	imes 10^{-6} \,F$. The new charge on the capacitor is $q = CV = (80 	imes 10^{-6}) 	imes 30 = 2400 	imes 10^{-6} \,C = 2.4 	imes 10^{-3} \,C$.The charge that passes through the wire is $\Delta q = q' - q = 48 	imes 10^{-3} - 2.4 	imes 10^{-3} = 45.6 	imes 10^{-3} \,C$.

Similar Questions

Between the plates of a parallel plate capacitor,a plate of thickness $t_1$ and dielectric constant $k_1$ is placed. In the rest of the space,there is another plate of thickness $t_2$ and dielectric constant $k_2$. The potential difference across the capacitor will be

$A$ parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$. What will be the capacitance if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

$A$ capacitor, when filled with a dielectric $K = 3$, has charge $Q_0$, voltage $V_0$, and electric field $E_0$. If the dielectric is replaced with another one having $K = 9$, the new values of charge, voltage, and field will be respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module