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Question

(D) The capacitor can be considered as two capacitors in parallel,one with dielectric and one without.Area of dielectric part $A_1 = A/3$,area of air

Vedclass · Accepted Answer

$(A,D)$

Vedclass · Answer

(D) The capacitor can be considered as two capacitors in parallel,one with dielectric and one without.Area of dielectric part $A_1 = A/3$,area of air part $A_2 = 2A/3$.Capacitance with dielectric: $C_1 = \frac{K \varepsilon_0 A}{3d}$.Capacitance without dielectric: $C_2 = \frac{\varepsilon_0 (2A/3)}{d} = \frac{2 \varepsilon_0 A}{3d}$.Total capacitance $C = C_1 + C_2 = \frac{K \varepsilon_0 A}{3d} + \frac{2 \varepsilon_0 A}{3d} = \frac{(K+2) \varepsilon_0 A}{3d}$.Ratio $\frac{C}{C_1} = \frac{(K+2) \varepsilon_0 A / 3d}{K \varepsilon_0 A / 3d} = \frac{K+2}{K}$. Thus,$(D)$ is correct.Since the plates are connected to the same potential difference $V$,the electric field in both parts is $E_1 = E_2 = V/d$. Thus,$\frac{E_1}{E_2} = 1$. Thus,$(A)$ is correct.Charges are $Q_1 = C_1 V$ and $Q_2 = C_2 V$. Ratio $\frac{Q_1}{Q_2} = \frac{C_1}{C_2} = \frac{K \varepsilon_0 A / 3d}{2 \varepsilon_0 A / 3d} = \frac{K}{2}$. Thus,$(C)$ is incorrect.The correct options are $(A)$ and $(D)$.

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