A slab of material of dielectric constant K h | vedclass

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

Question

(A) The capacitance of a parallel plate capacitor without a dielectric is $C_0 = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t = (4/5

Vedclass · Accepted Answer

$\frac{5K}{K + 4}$

Vedclass · Answer

(A) The capacitance of a parallel plate capacitor without a dielectric is $C_0 = \frac{\epsilon_0 A}{d}$.When a dielectric slab of thickness $t = (4/5)d$ and dielectric constant $K$ is inserted, the new capacitance $C$ is given by the formula $C = \frac{\epsilon_0 A}{d - t + (t/K)}$.Substituting $t = (4/5)d$ into the formula:$C = \frac{\epsilon_0 A}{d - (4/5)d + ((4/5)d/K)}$$C = \frac{\epsilon_0 A}{(1/5)d + (4d/5K)}$$C = \frac{\epsilon_0 A}{(d/5) [1 + (4/K)]} = \frac{\epsilon_0 A}{(d/5) [(K + 4)/K]}$$C = \frac{5K \epsilon_0 A}{d(K + 4)}$Since $C_0 = \frac{\epsilon_0 A}{d}$, we have $C = C_0 \cdot \frac{5K}{K + 4}$.Therefore, the ratio $(C/C_0) = \frac{5K}{K + 4}$.

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel plate capacitor but has a thickness $(4/5)d$, where $d$ is the separation of the plates. The capacitance in the presence and absence of dielectric are $C$ and $C_0$ respectively. The ratio $(C/C_0)$ is

Similar Questions

$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module