New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Capacitor question

2021 · 25 Jul · Shift 2 · Q54

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Physics
Capacitor
2021 · 25 Jul · Shift 2 · Q54

Capacitor question

2021 · 25 Jul · Shift 2 · Q54

JEE MainPhysicsCapacitorMCQ+4 / −1

If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as :

A
${q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)$
B
${q_b} = {q_f}\left( {1 - {1 \over k}} \right)$
C
${q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)$
D
${q_b} = {q_f}\left( {1 + {1 \over k}} \right)$

View written solutionFree

Correct answer: B

Relation between free charge and bound charge in a dielectric-filled capacitor

When a dielectric of dielectric constant $k$ is placed between the plates of a capacitor, polarization occurs.

Free charge on capacitor plates = $q_f$
Bound charge induced on dielectric surfaces = $q_b$

The electric field inside the dielectric is reduced due to polarization.

Use polarization relation

For a linear dielectric,

\vec D = \varepsilon_0 \vec E + \vec P

and also

\vec D = \varepsilon \vec E = k\varepsilon_0 \vec E

Hence,

\vec P = \vec D - \varepsilon_0 \vec E

But since

\vec D = k\varepsilon_0 \vec E,

we get

\vec P = k\varepsilon_0 \vec E - \varepsilon_0 \vec E = (k-1)\varepsilon_0 E

Now,

D = \frac{q_f}{A}, \qquad P = \frac{q_b}{A}

where $A$ is plate area.

So,

\frac{q_b}{A} = (k-1)\varepsilon_0 E

and

\frac{q_f}{A} = k\varepsilon_0 E

Divide the two equations:

\frac{q_b}{q_f} = \frac{k-1}{k}

Therefore,

q_b = q_f\left(1-\frac{1}{k}\right)

Check options

A: $q_f\left(1-\frac{1}{\sqrt{k}}\right)$ ❌
B: $q_f\left(1-\frac{1}{k}\right)$ ✅
C: $q_f\left(1+\frac{1}{\sqrt{k}}\right)$ ❌
D: $q_f\left(1+\frac{1}{k}\right)$ ❌

Thus the correct option is B.

More from Capacitor

The material filled between the plates of a parallel plate capacitor has resistivity 200 $Ω$ m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then…2021 · MCQ
A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the… Includes diagram2021 · MCQ
Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same connected in series. Calculate the ratio of…2021 · MCQ
Calculate the amount of charge on capacitor of 4 $μ$ F. The internal resistance of battery is 1 $Ω$ : Includes diagram2021 · MCQ
Three capacitors C1 = 2 $μ$ F, C2 = 6 $μ$ F and C3 = 12 $μ$ F are connected as shown in figure. Find the ratio of the charges on capacitors C1, C2 and C3 respectively : Includes diagram2021 · MCQ
A capacitor of capacitance C = 1 $μ$ F is suddenly connected to a battery of 100 volt through a resistance R = 100 $Ω$ . The time taken for the capacitor to be charged to get 50 V is : [Take ln 2 = 0.69] Includes diagram2021 · MCQ
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be : (Given area of plate = A) Includes diagram2021 · MCQ
Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across…2021 · MCQ