A 60 mu F parallel plate capacitor whose pla | vedclass

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

Question

(B) Initial capacitance $C = 60 \ \mu F$. Initial potential $V = 250 \ V$. Charge $q = CV = 60 \ \mu F 	imes 250 \ V = 15000 \ \mu C$.When a dielectr

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) Initial capacitance $C = 60 \ \mu F$. Initial potential $V = 250 \ V$. Charge $q = CV = 60 \ \mu F 	imes 250 \ V = 15000 \ \mu C$.When a dielectric slab of thickness $t = 3 \ mm$ and dielectric constant $K = 5$ is inserted between plates separated by $d = 6 \ mm$,the new capacitance $C'$ is given by $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.Since $C = \frac{\epsilon_0 A}{d}$,we have $C' = C \left[ \frac{d}{d - t + \frac{t}{K}} ight] = 60 \ \mu F \left[ \frac{6}{6 - 3 + \frac{3}{5}} ight] = 60 \ \mu F \left[ \frac{6}{3 + 0.6} ight] = 60 \ \mu F \left[ \frac{6}{3.6} ight] = 60 \ \mu F 	imes \frac{60}{36} = 60 \ \mu F 	imes \frac{5}{3} = 100 \ \mu F$.Since the charging source is removed,the charge $q$ remains constant. Thus,$q = C'V' \Rightarrow 15000 \ \mu C = 100 \ \mu F 	imes V' \Rightarrow V' = 150 \ V$.The change in potential difference is $\Delta V = V - V' = 250 \ V - 150 \ V = 100 \ V$.

Similar Questions

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

$A$ parallel plate capacitor has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?

$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

Vedclass Test Series

Exam Paper Generator

Online Exam Module