A parallel plate capacitor with air as dielec | vedclass

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

Question

(D) Initial capacitance with air $C_0 = 4 \mu F$. After filling with dielectric $(K = 5)$,the new capacitance is $C = K C_0 = 5 	imes 4 \mu F = 20 \m

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(D) Initial capacitance with air $C_0 = 4 \mu F$. After filling with dielectric $(K = 5)$,the new capacitance is $C = K C_0 = 5 	imes 4 \mu F = 20 \mu F$. The charge on the capacitor is $Q = C V = 20 \mu F 	imes 100 \ V = 2000 \mu C = 2 	imes 10^{-3} \ C$. Since the capacitor is disconnected from the battery,the charge $Q$ remains constant. Initial energy stored with dielectric: $U_i = \frac{Q^2}{2C} = \frac{(2 	imes 10^{-3})^2}{2 	imes 20 	imes 10^{-6}} = \frac{4 	imes 10^{-6}}{40 	imes 10^{-6}} = 0.1 \ J$. Final capacitance after removing the dielectric: $C_f = C_0 = 4 \mu F$. Final energy stored: $U_f = \frac{Q^2}{2C_f} = \frac{(2 	imes 10^{-3})^2}{2 	imes 4 	imes 10^{-6}} = \frac{4 	imes 10^{-6}}{8 	imes 10^{-6}} = 0.5 \ J$. Work done by external agent $W = U_f - U_i = 0.5 \ J - 0.1 \ J = 0.4 \ J$.

Similar Questions

$A$ parallel plate capacitor has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.

Explain the effect of a dielectric on the capacitance of a parallel plate capacitor and obtain the formula for the dielectric constant.

$A$ capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process

Vedclass Test Series

Exam Paper Generator

Online Exam Module