A slab of material with a dielectric constant | vedclass

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

Question

(C) In the absence of a dielectric,the electric field between the two plates is $E_0 = \frac{V_0}{d}$.When the dielectric slab is inserted,the electri

Vedclass · Accepted Answer

$\frac{V_0}{2}$

Vedclass · Answer

(C) In the absence of a dielectric,the electric field between the two plates is $E_0 = \frac{V_0}{d}$.When the dielectric slab is inserted,the electric field inside the dielectric becomes $E = \frac{E_0}{K}$,where $K=3$ is the dielectric constant.The thickness of the dielectric slab is $t = \frac{3}{4}d$,and the thickness of the air gap is $d - t = d - \frac{3}{4}d = \frac{1}{4}d$.The new potential difference $V$ is the sum of the potential drops across the air gap and the dielectric slab:$V = E_0 \left(\frac{1}{4}dight) + E \left(\frac{3}{4}dight)$$V = E_0 \left(\frac{1}{4}dight) + \frac{E_0}{K} \left(\frac{3}{4}dight)$$V = E_0 d \left[ \frac{1}{4} + \frac{3}{4K} ight]$Since $E_0 d = V_0$ and $K = 3$:$V = V_0 \left[ \frac{1}{4} + \frac{3}{4 	imes 3} ight]$$V = V_0 \left[ \frac{1}{4} + \frac{1}{4} ight]$$V = V_0 \left( \frac{2}{4} ight) = \frac{V_0}{2}$

Similar Questions

$A$ parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$,its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is

$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

Two parallel plate air capacitors of the same capacity $C$ are connected in series to a battery of emf $E$. Then one of the capacitors is completely filled with a dielectric material of constant $K$. The change in the effective capacitance of the series combination is

The function of a dielectric in a capacitor is

$A$ parallel plate capacitor with air as the dielectric has capacitance $C$. $A$ slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module