A parallel plate capacitor has plates of area 'A' separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as \$k(x)=K(1+\alpha x)\$ where 'x' is the distance measured from one of the plates. If \$(\alpha d)<<1\$, the total capacitance of the system is best by the expression:

Refer image 1.Let \$k(x)=k(1+\alpha x)\$be the dielectric constantThe capacitor can be considered to be a series combination of a large number of capacitors of thickness \$\Delta x\$.Refer image 2.\$\displaystyle \dfrac{1}{C_{eq}}=\sum\dfrac{\Delta x}{k(x)A\epsilon_0}=\int^d_0\dfrac{dx}{k(x)A\epsilon_0}\$where \$\Delta x\rightarrow 0\$\$\displaystyle \dfrac{1}{C_{eq}}=\int^d_0\dfrac{dx}{k(1+\alpha x)A\epsilon_0}=\dfrac{1}{KA\epsilon_0}\int^d_0\dfrac{dx}{1+\alpha x}\$ \$=\dfrac{1}{KA\epsilon_0}\log_e(\dfrac{1+\alpha d}{1})\times \dfrac{1}{\alpha}\$\$=\dfrac{1}{K\alpha A\epsilon_0}\log_e(1+\alpha d)\$Now, \$\log (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-------\$\$\therefore \log_e(1+\alpha d)\simeq \alpha d-\dfrac{\alpha^2d^2}{2}\$\$\therefore \dfrac{1}{C_{eq}}=\dfrac{1}{k\alpha A\epsilon_e}\log_e(1+\alpha d)\$\$=\dfrac{1}{k\alpha A\epsilon_0}[\alpha d-\dfrac{\alpha^2d^2}{2}+-------]\$\$\simeq \dfrac{\alpha d}{k\alpha A\epsilon_0}[1-\dfrac{\alpha d}{2}]\simeq \dfrac{d}{KA\epsilon_0}[1-\dfrac{\alpha d}{2}]\$\$C_{eq}=\dfrac{KA\epsilon_0}{d[1-\dfrac{\alpha d}{2}]}=\dfrac{KA\epsilon_0}{d}[1-\dfrac{\alpha d}{2}]^{-1}\$Now, \$(1-x)^{-1}\simeq 1+x\$ \$[\because x<<1]\$\$\therefore C_{eq}=\dfrac{KA\epsilon_0}{d}[1+\dfrac{\alpha d}{2}]\$option (C) is correct.

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>>Class 12
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>> $Effects of Dielectrics in Capacitors$

Effects of Dielectrics in Capacitors

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Basic Knowledge of Effect of Dielectric on Capacitance of Parallel Plate Capacitor

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Q: A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of (i) capacitance.(ii) potential difference between the plates.(iii) electric field between the plates.(iv) the energy stored in the capacitor.

\$(i)\$ The capacitance increases as the dielectric constant \$K > 1\$. \$(ii)\$ Potential difference \$V =\dfrac{Q}{C}\$. As \$C\$ increases and \$Q\$ remains the same since the battery is disconnected, the p.d. between the plates decreases.\$(iii)\$ Electric field \$E =\dfrac{V}{d}\$ where \$V\$ is the p.d. and \$d\$ the separation between the plates. As \$V\$ decreases and d remains the same, electric field also decreases.\$(iv)\$ Energy stored in a capacitor \$U =\dfrac{1}{2}\dfrac{{Q}^{2}}{C}\$. As \$Q\$ is constant and \$C\$ increases, \$U\$ decreases.

Q: A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =\$ 10^{-12}\$ F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

\$\textbf{Step 1: Initial capacitance} \$Let the initial distance between the plates be \$d.\$Let \$A\$ be the area of each plate.Capacitance of the parallel plate capacitor is given by: \$C_i = \dfrac{A\in_0}{d} \$ \$....(1)\$\$\textbf{Step 2: Capacitance after inserting dielectric} \$Let \$k\$ be the dielectric constant. \$C_f = \dfrac{kA \in_0}{d'}\$As the distance between plates is reduced to half, new distance is \$d' = \dfrac{d}{2}\$ \$\therefore C_f = \dfrac{2kA\in_0}{d}= 2kC_i= 2\times 6\times 8pF= 96pF\$

Q: The capacitance of a parallel plate capacitor with air as medium is \$6 \mu F\$. With the introduction of a dielectric medium, the capacitance becomes \$30\mu F\$. The permittivity of the medium is :\$(\in_0= 8.85 \times 10^{-12} C^2N^{-1} m^{-2})\$

Capacitance of air capacitor \$C_0 = \dfrac{\varepsilon_0A}{d} = 6 \mu F\$ ......... (i)When a dielectric of permittivity \$\varepsilon_r\$ and dielectric constant K is introduced between the plates, thenCapacitance, \$C = \dfrac{K \varepsilon_0 A}{d} = 30 \mu F\$ ..... (ii)Dividing eq. (ii) by (i), we get\$\dfrac{C}{C_0} = \dfrac{\dfrac{K\epsilon_o A}{ d}}{\dfrac{\varepsilon_0 A}{d}} = \dfrac{30}{6}\$\$\Rightarrow K = 5\$\$\therefore\$ permittivity of the medium\$\varepsilon_0 = \varepsilon_0 K\$\$= 8.85 \times 10^{-12} \times 5 = 0.44 \times 10^{-10}\$

Q: A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness \$d_{1}\$ and dielectric constant \$K_{1}\$ and the other has thickness \$d_{2}\$ and dielectric constant \$K_{2}\$ as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (=\$d_{1}+d_{2}\$) and effective dielectric constant K. The K is

The capacities of two individual condensers are\${ C }_{ 1 }=\cfrac { { K }_{ 1 }{ \varepsilon }_{ 0 }A }{ { d }_{ 1 } } ;{ C }_{ 2 }=\cfrac { { K }_{ 2 }{ \varepsilon }_{ 0 }A }{ { d }_{ 2 } } \$The arrangement is equivalent to two capacitors joined in series\$\therefore\$ Equivalent capacitance, \$\cfrac { 1 }{ { C }_{ eq } } =\cfrac { 1 }{ { C }_{ 1 } } +\cfrac { 1 }{ { C }_{ 2 } } =\cfrac { { d }_{ 1 } }{ { K }_{ 1 }{ \varepsilon }_{ 0 }A } +\cfrac { { d }_{ 2 } }{ { K }_{ 2 }{ \varepsilon }_{ 0 }A } \$\$=\cfrac { 1 }{ { \varepsilon }_{ 0 }A } \left[ \cfrac { { d }_{ 1 } }{ { K }_{ 1 } } +\cfrac { { d }_{ 2 } }{ { K }_{ 2 } } \right] =\cfrac { 1 }{ { \varepsilon }_{ 0 }A } \left[ \cfrac { { { K }_{ 2 }d }_{ 1 }+{ K }_{ 1 }{ d }_{ 2 } }{ { K }_{ 1 }+{ K }_{ 2 } } \right] \$or \${ C }_{ eq }={ \varepsilon }_{ 0 }A\left( \cfrac { { K }_{ 1 }+{ K }_{ 2 } }{ { { K }_{ 2 }d }_{ 1 }+{ K }_{ 1 }{ d }_{ 2 } } \right) ....(i)\quad \$Also \${ C }_{ eq }=\cfrac { K{ \varepsilon }_{ 0 }A }{ { d }_{ 1 }+{ d }_{ 2 } } \$From (i) and (ii) \$\quad \quad { \varepsilon }_{ 0 }A\left( \cfrac { { K }_{ 1 }+{ K }_{ 2 } }{ { { K }_{ 2 }d }_{ 1 }+{ K }_{ 1 }{ d }_{ 2 } } \right) ={ \varepsilon }_{ 0 }A\left( \cfrac { K }{ { d }_{ 1 }+{ d }_{ 2 } } \right) \$\$\therefore K=\cfrac { { K }_{ 1 }{ K }_{ 2 }\left( { d }_{ 1 }+{ d }_{ 2 } \right) }{ { K }_{ 1 }{ d }_{ 2 }+{ { K }_{ 2 }d }_{ 1 } } \$

Q: A parallel plate capacitor is of area \$6\ cm^{2}\$ and separation \$3\ mm\$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant \$K_{1}, = 10\$, \$K_{2} = 12\$ and,\$K_{3}=14\$. The dielectric constant of a material which when fully inserted in above capacitor gives same capacitance would be

Let the dielectric constant of material used be \$K\$.\$C_1+C_2+C_2+C_3=C_{eq}\$\$\therefore \dfrac {10\epsilon_{0} A/3}{d} + \dfrac {12\epsilon_{0}A/3}{d} + \dfrac {14\epsilon_{0}A/3}{d} = \dfrac {K\epsilon_{0}A}{d}\$On solving \$\Rightarrow K = 12\$.

Q: A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness \$3d/4\$, where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

\$C = \cfrac{A \epsilon_0}{d}\$\$C' = \cfrac{A \epsilon_0}{d - t + \cfrac{t}{k}}\$Put \$t = \cfrac{3d}{4}\$ ;\$C' = \cfrac{4k}{3 + k} . \cfrac{A \epsilon_0}{d}\$\$C' = \cfrac{4k}{3 + k} . C\$

Q: A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

Capacitance \$C=\varepsilon\dfrac{A}{d}\$ , \$A-area\ of\ the\ capacitor\ ; d-distance\ between\ the\ plates\$For a charged capacitor \$Q=CV\$As the distance between the plates of the capacitor is increased the capacitance \$C\$ decreases \$\ \ \ (\because C \propto \dfrac{1}{d})\$ As capacitance decreases the potential difference between the plates \$V \ increases\$

Q: A parallel plate capacitor with air between the plates has capacitance of \$9 pF\$. The separation between its plates is '\$d\$'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant \$k_1=3\$ and thickness \$\dfrac {d}{3}\$ while the other one has dielectric constant \$k_2=6\$ and thickness \$\dfrac {2d}{3}\$. Capacitance of the capacitor is now :

The air filled capacitance , \$C=\dfrac{A\epsilon_0}{d}=9 pF\$According to question there are two capacitor with capacitances \$C_1 \$ and \$ C_2\$ . and they are in series .Here, \$C_1=\dfrac{Ak_1\epsilon_0}{(d/3)}=9 \dfrac{A\epsilon_0}{d}=9C \$ as \$k_1=3\$and \$C_2= \dfrac{Ak_2\epsilon_0}{(2d/3)}=9 \dfrac{A\epsilon_0}{d}=9C \$ as \$k_2=6\$The equivalent capacitance, \$C_{eq}=\dfrac{C_1C_2}{C_1+C_2}=\dfrac{(9C)(9C)}{9C+9C}=\dfrac{9}{2}C=(9/2) 9=40.5 pF\$

Q: A parallel plate capacitor has plates of area 'A' separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as \$k(x)=K(1+\alpha x)\$ where 'x' is the distance measured from one of the plates. If \$(\alpha d)<<1\$, the total capacitance of the system is best by the expression:

Refer image 1.Let \$k(x)=k(1+\alpha x)\$be the dielectric constantThe capacitor can be considered to be a series combination of a large number of capacitors of thickness \$\Delta x\$.Refer image 2.\$\displaystyle \dfrac{1}{C_{eq}}=\sum\dfrac{\Delta x}{k(x)A\epsilon_0}=\int^d_0\dfrac{dx}{k(x)A\epsilon_0}\$where \$\Delta x\rightarrow 0\$\$\displaystyle \dfrac{1}{C_{eq}}=\int^d_0\dfrac{dx}{k(1+\alpha x)A\epsilon_0}=\dfrac{1}{KA\epsilon_0}\int^d_0\dfrac{dx}{1+\alpha x}\$ \$=\dfrac{1}{KA\epsilon_0}\log_e(\dfrac{1+\alpha d}{1})\times \dfrac{1}{\alpha}\$\$=\dfrac{1}{K\alpha A\epsilon_0}\log_e(1+\alpha d)\$Now, \$\log (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-------\$\$\therefore \log_e(1+\alpha d)\simeq \alpha d-\dfrac{\alpha^2d^2}{2}\$\$\therefore \dfrac{1}{C_{eq}}=\dfrac{1}{k\alpha A\epsilon_e}\log_e(1+\alpha d)\$\$=\dfrac{1}{k\alpha A\epsilon_0}[\alpha d-\dfrac{\alpha^2d^2}{2}+-------]\$\$\simeq \dfrac{\alpha d}{k\alpha A\epsilon_0}[1-\dfrac{\alpha d}{2}]\simeq \dfrac{d}{KA\epsilon_0}[1-\dfrac{\alpha d}{2}]\$\$C_{eq}=\dfrac{KA\epsilon_0}{d[1-\dfrac{\alpha d}{2}]}=\dfrac{KA\epsilon_0}{d}[1-\dfrac{\alpha d}{2}]^{-1}\$Now, \$(1-x)^{-1}\simeq 1+x\$ \$[\because x<<1]\$\$\therefore C_{eq}=\dfrac{KA\epsilon_0}{d}[1+\dfrac{\alpha d}{2}]\$option (C) is correct.

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Capacitance of the Parallel Plate Capacitor with Dielectric Slab

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Important Questions

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of
(i) capacitance.
(ii) potential difference between the plates.
(iii) electric field between the plates.
(iv) the energy stored in the capacitor.

Hard

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A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $K_{1}, K_{2}, K_{3}$ as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor then its dielectric constant K is given by

Medium

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A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = $1 0^{- 12}$ F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Medium

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The capacitance of a parallel plate capacitor with air as medium is $6 μ F$ . With the introduction of a dielectric medium, the capacitance becomes $30 μ F$ . The permittivity of the medium is :
$(\in_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2})$

Medium

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A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_{1}$ and dielectric constant $K_{1}$ and the other has thickness $d_{2}$ and dielectric constant $K_{2}$ as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (= $d_{1} + d_{2}$ ) and effective dielectric constant K. The K is

Hard

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A parallel plate capacitor is of area $6 c m^{2}$ and separation $3 m m$ . The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant $K_{1}, = 10$ , $K_{2} = 12$ and, $K_{3} = 14$ . The dielectric constant of a material which when fully inserted in above capacitor gives same capacitance would be

Medium

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A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness $3 d / 4$ , where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

Medium

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A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

Medium

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A parallel plate capacitor with air between the plates has capacitance of $9 p F$ . The separation between its plates is ' $d$ '. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_{1} = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $k_{2} = 6$ and thickness $\frac{2 d}{3}$ . Capacitance of the capacitor is now :

Hard

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A parallel plate capacitor has plates of area 'A' separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as $k (x) = K (1 + α x)$ where 'x' is the distance measured from one of the plates. If $(α d) < < 1$ , the total capacitance of the system is best by the expression:

Hard

JEE Mains

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Revise with Concepts

Basic Knowledge of Effect of Dielectric on Capacitance of Parallel Plate Capacitor

Electric Displacement

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Quick Summary With Stories

Capacitance of the Parallel Plate Capacitor with Dielectric Slab

Relation between dielectric constant and suspectibility, Dielectric strength

Electric Displacement

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

The capacitance of a parallel plate capacitor with air as medium is 6μF. With the introduction of a dielectric medium, the capacitance becomes 30μF. The permittivity of the medium is : (∈0​=8.85×10−12C2N−1m−2)

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness 3d/4, where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = $1 0^{- 12}$ F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

The capacitance of a parallel plate capacitor with air as medium is $6 μ F$ . With the introduction of a dielectric medium, the capacitance becomes $30 μ F$ . The permittivity of the medium is :
$(\in_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2})$

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness $3 d / 4$ , where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?