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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>>Electrostatic Potential and Capacitance
>> $Effects of Dielectrics in Capacitors$

Effects of Dielectrics in Capacitors

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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 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: Derive the expression for the capacitance of a parallel plate capacitor.

Let d be the separation of two conducting parallel plates , each of area A .If charge q is given to first plate , then charge -q is induced on inner surface of second plate and charge +q induced on the outer surface , is earthed .Now , the electric field between the two plates will be , E=V/d or V=Ed ...........eq1If \sigma is he surface density of the plates , then the electric field between two plates is given by , E=\sigma/\varepsilon_{0} ...............eq2where \varepsilon_{0}= absolute permittivity of the free space.Putting the value of E from eq2 to eq1 , we get V=\frac{\sigma}{\varepsilon_{0}}d ................eq3but we have , \sigma=q/A hence , V=\frac{qd}{\varepsilon_{0}A} Now , capacitance of parallel capacitor , C=\frac{q}{qd/\varepsilon_{0}A} or C=\frac{\varepsilon_{0}A}{d}

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.

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: A parallel plate capacitor is charged by a battery.The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates. How will (i) its capacitance, (ii) electric field between the plates and (iii) energy stored in the capacitor be affected ? Justify your answer giving necessary mathematical expressions for each case.

The capacitance without dielectric is C=\dfrac{A\epsilon_0}{d} When dielectric slab is inserted, the capacitance becomes, C'=\dfrac{AK\epsilon_0}{d}=KC where K be the dielectric constant. i)Thus, the capacitance will increase K times of the initial.ii) As the battery is disconnected so the charge on capacitor remains constant.Since, Q=CV so potential V will decrease and also E=V/d so the field E will also decrease.iii) Stored energy, U=\dfrac{Q^2}{2C} . As charge Q is constant and C is increasing so energy will decrease.

Q: The potential difference between the two plates of a parallel plate capacitor is constant. When air between the plates is replaced by dielectric material, the electric field intensity :

In general capacitance of parallel plate capacitor is given by: C=\dfrac{k\epsilon_0 A}{d} where C is capacitance, k is relative permittivity of dielectric material, \epsilon_0 is permittivity of free space constant, A is area of plates and d is distance between them. Therefore, the capacitance of parallel plates is increased by the insertion of a dielectric material. Further, the capacitance is inversely proportional to the electric field between the plates, and hence the presence of the dielectric decreases the effective electric field.

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

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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 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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Derive the expression for the capacitance of a parallel plate capacitor.

Medium

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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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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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A parallel plate capacitor is charged by a battery.
The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.
How will (i) its capacitance, (ii) electric field between the plates and (iii) energy stored in the capacitor be affected ? Justify your answer giving necessary mathematical expressions for each case.

Medium

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Predict the polarity of the capacitor in the situation described by figure

Medium

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The potential difference between the two plates of a parallel plate capacitor is constant. When air between the plates is replaced by dielectric material, the electric field intensity :

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

View solution

Revise with Concepts

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

Electric Displacement

Learn with Videos

Quick Summary With Stories

Capacitance of the Parallel Plate Capacitor with Dielectric Slab

Relation between dielectric constant and suspectibility, Dielectric strength

Electric Displacement

Important Questions

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?

Derive the expression for the capacitance of a parallel plate capacitor.

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?

Predict the polarity of the capacitor in the situation described by figure

The potential difference between the two plates of a parallel plate capacitor is constant. When air between the plates is replaced by dielectric material, the electric field intensity :

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 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?

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?