Assertion is incorrect but Reason is correct

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 12

A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 3d$ . The slab is equidistant from the plates. The capacitor is given some charge. As $x$ goes from $0$ to $3d$ :

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the magnitude of the electric field remains the same
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the direction of the electric field remains the same
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the electric potential increases continuously
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the electric potential increases at first, then decreases and again increases

Explanation

$x$ from

$0-3d$

1) Magnitude of

$\overset { \rightarrow }{ E }$ changes due to dielectric slab. In region A and C is same but not in B

2) the direction of

$\overset { \rightarrow }{ E }$ remains same throughout in all regions A,B,C

3) Electric potential also changes, since electric field varies, so potential also changes.

4) Electric potential first increases and then decreases due to dielectric slab and then again increases.

A parallel plate capacitor of capacity

$5 \mu F$ and plate separation

$6 cm$ is connected to a

$1V$ battery and is charged. A dielectric of dielectric constant

$4$ and thickness

$4 cm$ is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is:

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2 $\mu$ C
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3 $\mu$ C
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5 $\mu$ C
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10 $\mu$ C

Three charges

$Q,+q$ and

$+q$ are placed at the vertices of a right angle isosceles triangle as shown. The net electrostatic energy of the configuration is zero, if

$Q$ is equal to :

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$\dfrac { -q }{ 1+\sqrt { 2 } }$
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$\dfrac { -2q }{ 2+\sqrt { 2 } }$
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$-2q$
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$+q$

Explanation

As triangle is isosceles right angle, so

$AB=BC=a$ and

$AC=\sqrt{2}a$

The net electrostatic energy of the system is

$U=U_{AB}+U_{BC}+U_{AC}=0$

$\dfrac{1}{4\pi\epsilon_0}(\dfrac{Qq}{a}+\dfrac{q^2}{a}+\dfrac{Qq}{\sqrt{2}a})=0$

$Q(1+\dfrac{1}{\sqrt{2}})=-q$

$Q=\dfrac{-\sqrt{2}q}{\sqrt{2}+1}=\dfrac{-2q}{2+\sqrt 2}$

Three charges $Q$ , $+q$ and $+q$ are placed at the vertices of a right angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if $Q$ is equal to :

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$\dfrac{-q}{1+\sqrt{2}}$
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$\dfrac{-\sqrt{2}q}{1+\sqrt{2}}$
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$-2q$
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$+q$

Explanation

We know P.E between two charges

$=\dfrac {k q_1 q_2}{r.}$

Now total P. E

$=P\cdot E_{AB}+P\cdot E_{AC}+P\cdot E_{BC}.$

$P\cdot E_{ta}=\dfrac {k Q q}{\sqrt 2 a}+\dfrac {k Q q}{a}+\dfrac {k q^2}{a}$

$0=\dfrac {k q}{a}(\dfrac {Q}{\sqrt 2}+Q+q)$

$\Rightarrow 0=Q(1+\dfrac {1}{\sqrt 2})+q$

$\Rightarrow -q=Q(\dfrac {\sqrt 2+1}{\sqrt 2})$

$\Rightarrow Q=\dfrac {-\sqrt 2 q}{\sqrt 2+ 1}$

An alpha particle of $5 MeV$ at a large distance proceeds towards a gold nucleus $(Z=79)$ to make a head on collision. The closest distance of approach from the centre of gold nucleus is:

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$20 fm$
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$15 fm$
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$10 fm$
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$45 fm$

Explanation

Initial energy of

$\alpha\text{-particle} =V\times e=5\times 10^6\times (1.6\times 10^{-19})J.$
Let

$x$ be the closest distance with nucleus.
Now closest distance is when the energy of

$\alpha$ particle is totally converted to potential energy between them.

$\therefore 5\times 10^6\times1.6\times 10^{-19}=\dfrac{kq_1q_2}{r}=\dfrac {k(79\times e)(2\times e)}{x.}$

$\Rightarrow x=\dfrac {(9\times 10^9)(79\times 1\cdot 6\times 10^{-19})(2\times 1\cdot 6\times 10^{-19})}{5\times 10^6\times 1.6\times 10^{-19}}$

$\Rightarrow x=45\times 10^{-15} m$

Figure shows a hemisphere of charge Q and radius R and a sphere of charge 2Q and radius R. The total potential energy of hemisphere is

$U_{H}$ and that the sphere is

$U_{S}$ . Then,

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$2U_{H}=U_{S}$
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$2U_{H} < U_{S}$
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$2U_H > U_S$
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$U_{H}=U_{S}$

Explanation

As both hemispherical sphere and sphere have charge so there are some interaction energy in between them. here,

$U_S=2U_H+$ interaction energy term. Thus,

$U_S>2U_H$ .

An isolated parallel plate capacitor is charged upto a certain potential difference. When a

$3mm$ thick slab is introduced between the plates then in order to maintain the same potential difference, the distance between the plates is increased by

$2.4mm$ . Find the dielectric constant of the slab. (Assume charge remains constant)

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5
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10
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2.5
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7.5

Explanation

$C_f=\dfrac {\varepsilon_0A}{d'-t(1-\frac {1}{K})}$

$C_i=C_f$

$\dfrac {\varepsilon_0A}{d}=\dfrac {\varepsilon_0A}{d'-t(1-\dfrac {1}{K})}\Rightarrow d=d'-t(1-\dfrac {1}{K})$

$d'=d-2.4\times 10^{ -3 }$

$t=3mm$

$d=d+(2\cdot 4\times 10^{-3})-3\times 10^{-3}(1-\dfrac {1}{K})$

$K=5$

Three identical square metal plates

$M_1, M_2$ and

$M_3$ of side 10 cm & 5 mm thick are arranged as shown in figure. The plates are separated by sheets of paper 0.5 mm thick & of dielectric constantThe outer plates are connected together & connected to lower potential while inner plate to higher potential terminal of a battery. Capacitance between the terminals of the battery is :

$(\epsilon _{0} = 8.9 10^{-12} C^{2}/N-m^{2})$

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$1780pF$
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$890 pF$
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$445 pF$
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$222.5 pF$

Explanation

Here three plates will make two capacitors with capacitance

$C=\dfrac{KA\epsilon_0}{d}$

As the battery terminals for two capacitors are common so they are in parallel.

Thus, the equivalent capacitance across battery is:

$C_{eq}=C+C=2C=2\times \dfrac{KA\epsilon_0}{d}=2\times \dfrac{5(0.1)^2(8.9\times 10^{12})}{0.5\times 10^{-3}}=1780\times 10^{-12} F=1780 pF$

In the figure shown, a parallel plate capacitor has a dielectric of width

$d/2$ and dielectric constant

$K = 2.$ The other dimensions of the dielectric are same as that of the plates. The plates

$P_1$ and

$P_2$ of the capacitor have area 'A' each. The energy of the capacitor is :

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$\dfrac{\epsilon _{0}AV^{2}}{3d}$
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$\dfrac{2\epsilon _{0}AV^{2}}{d}$
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$\dfrac{3}{2}\dfrac{\epsilon _{0}AV^{2}}{d}$
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$\dfrac{2\epsilon _{0}AV^{2}}{3d}$

Explanation

Using

$Energy=\dfrac{1}{2} \times C_{equivalent}\times V^2$

We need to find the equivalent capacitance of the capacitor.

Using law for series combination of capacitors,

$\dfrac{1}{C_{net}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}$

where, $C_1 = \dfrac{A\epsilon_0}{x}$ ( x is the width between capacitor walls and dielectric on one side )

$C_2 = \dfrac {A\epsilon_0\kappa}{(\dfrac{d}{2})} = \dfrac{4A\epsilon_0}{d}$

$C_3 = \dfrac{A\epsilon_0}{(\dfrac{d}{2}-x)}$

therefore,

$\dfrac{1}{C_{net}} = \dfrac{x}{(A\epsilon_0)} + \dfrac{(\dfrac{d}{2}-x)}{(A\epsilon_0)} + \dfrac{d}{(4A\epsilon_0)} = \dfrac{3d}{(4A\epsilon_0)} \Rightarrow C_{net} = \dfrac{4A\epsilon_0}{3d}$

Therefore energy $= \dfrac{2A\epsilon_0 V^2}{3d}$

The capacity of a parallel plate condenser without any dielectric is C. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant

$3$ , the capacity of the condenser becomes:

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$\dfrac {3}{4}C$
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$\dfrac {9}{2}C$
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$\dfrac {2}{3}c$
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$\dfrac {3}{2}C$

Explanation

The capacitance of a parallel plate capacitor is given by

$C = \dfrac {\varepsilon_o \varepsilon_r A}{d}$ ,

where $A$ is the area of the plate and $d$ is the distance between them.

In the first case, there was no dielectric medium, i.e., $\varepsilon_r = 1$ .

So we have $C = \dfrac {\varepsilon_o A}{d}$

In the second case, the distance is doubled, i.e., $2d$ and a substance of dielectric constant $3$ is inserted between the plates.

Hence capacitance will change to

$C' = \dfrac {3 \varepsilon_o A}{2d}$

So, $\dfrac{C'}{C} = \dfrac {3}{2}$

$\implies C' = \dfrac {3}{2} C$

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct

Explanation

We know that inside the conductor electric field is zero. So,

$\vec E_{Net}=\vec E_{ind}+\vec E_{ext}=0$

$\Rightarrow \vec E_{ind}=-\vec E_{ext}=-\frac {Kq}{r^2}\hat i$
Thus, assertion is incorrect and reason is correct.
Ans:

$(D)$

A parallel plate capacitor is connected to a cell. It's positive plate A and the negative plate B have charges

$+Q$ and

$-Q$ respectively. A third plate C identical to A and B with charge

$+Q$ is now introduced midway between the plates A and B parallel to them What is the charge on the inner surface of A ?

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$Q$
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$\displaystyle \frac{\mathrm{Q}}{2}$
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$\displaystyle \frac{3\mathrm{Q}}{2}$
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$\displaystyle \frac{\mathrm{Q}}{4}$

Explanation

Let the capacitance of the A+B system be given by C =

$\dfrac{\epsilon A}{d}$
Now when the plate is introduced in middle, the distance of A+C and C+B system is changed to

$\dfrac{d}{2}$ So the capacitance will be

$\dfrac{2\epsilon A}{d}$ = 2C
Now, Total charge in the system is Q assume A+C has q charge and C+B has Q+q charge
So,

$\dfrac{q}{2C}$ +

$\dfrac{Q+q}{2C}$ =

$\dfrac{Q}{C}$
So, q =

$\dfrac{Q}{2}$

A parallel plate capacitor of area

$A$ and separation

$d$ is charged to potential difference

$V$ and removed from the charging source. A dielectric slab of constant

$K = 5$ , thickness

$d$ and area

$\frac {A}{3}$ is inserted, as shown in the figure. Let

$\sigma_1$ be free charge density at the conductor-dielectric surface and

$\sigma_2$ be the charge density at the conductor-vacuum surface :

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The electric field have the same value inside the dielectric as in the free space between the plates.
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The ratio $\dfrac {\sigma_1}{\sigma_2}$ is equal to $\dfrac {1}{5}$
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The new capacitance is $\dfrac {7\epsilon_0A}{3d}$
0%
The new potential difference is $\dfrac {3}{7}V$

Explanation

Since the capacitors are attached in parallel, the potential across the capacitors is same and so is the electric field inside the capacitors

$(Ed=V)$ .Therefore,

$let C = \dfrac{A\epsilon_0}{d}$

Now,

$\dfrac{q}{\dfrac{(2A\epsilon_0)}{(3d)} } = \dfrac{(CV-q)}{\dfrac{(A\kappa \epsilon_0}{(3d)} ) }$

$\dfrac{3q}{2C} = \dfrac{3(CV-q)}{5C} \Rightarrow 7q = 2CV$

Therefore

$V = \dfrac{q}{\dfrac{2A\epsilon_0}{(3d)} } = \dfrac{\dfrac{2CV}{7}}{\dfrac{2C}{3}}= \dfrac{3V}{7}$

Also

$C_{net} = \dfrac{2A\epsilon_0}{(3d) }+ \dfrac{A\kappa \epsilon_0}{(3d)} = \dfrac{7A\epsilon_0}{(3d) }$

Also ratio of charge densities,

$\dfrac{\sigma_2}{\sigma_1} = \dfrac{\dfrac{q}{\dfrac{2A}{3}}}{\dfrac{CV-q)}{\dfrac{A}{3}}} = \dfrac{\dfrac{\dfrac{2CV}{7}}{\dfrac{2A}{3}}}{\dfrac{\dfrac{5CV}{7}}{\dfrac{A}{3} }}= \dfrac{1}{5}$

A dielectric slab of thickness d is inserted in a parallel plate capacitor where negative plate is at

$x = 0$ and positive plate is at

$x = 3d$ . The slab is equidistant from the plates. The capacitor is given some charge. As x goes from

$0$ to

$3d$ .

0%
The magnitude of electric field remains the same.
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The direction of electric field remain the same.
0%
The electric potential increases continuously.
0%
The electric potential increase at first then decrease and again increase.

Explanation

After inserting dielectric slab electric field will change because inside slab

$E=\dfrac{E_0}{K}$ .where k is the dielectric constant.
but field direction will be same. As the potential is a continuous function of x, potential will increase continuously.
Ans:

$(B),(C)$

Two identical capacitors

$1$ and

$2$ are connected in series to a battery as shown in figure. Capacitor

$2$ contains a dielectric slab of dielectric constant

$k$ as shown.

$Q_1$ and

$Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are

$Q'_1$ and

$Q'_2$ .
Then

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$\displaystyle\frac{Q'_1}{Q_1} = \displaystyle\frac{k+1}{k}$
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$\displaystyle\frac{Q'_2}{Q_2} = \displaystyle\frac{k+1}{2}$
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$\displaystyle\frac{Q'_2}{Q_2} = \displaystyle\frac{k+1}{2k}$
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$\displaystyle\frac{Q'_1}{Q_1} = \displaystyle\frac{k}{2}$

Explanation

The capacitor 1 has capacitance

$C$ and as dielectric contains in capacitor 2 so its capacitance becomes

$kC$ .

The net capacitance

$C_{eq}=\dfrac{C.kC}{C+kC}=\dfrac{Ck}{1+k}$

and

$Q_{eq}=C_{eq}E$

When dielectric is reemoved from capacitor 2, its capacitance becomes

$C$ .

now net capacitance

$C'_{eq}=\dfrac{C.C}{C+C}=\dfrac{C}{2}$

and

$Q'_{eq}=C'_{eq}E=\dfrac{CE}{2}$

When the capacitors are connected in series, they each have the same charge as the net capacitance. Thus,

$Q_{eq}=Q_1=Q_2$ and

$Q'_{eq}=Q'_1=Q'_2$

$\therefore \dfrac{Q'_1}{Q_1}=\dfrac{Q'_2}{Q_2}=\dfrac{2k}{k+1}$

A point charge -q placed at the point A is

0%
In stable equilibrium along x-axis.
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In unstable equilibrium along y-axis.
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In stable equilibrium along y-axis.
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In unstable equilibrium along x-axis.

Explanation

If the potential energy of the system is minimum, it will be stable equilibrium, i.e,

$\frac{d^2U}{dx^2}>0$ and when potential energy is maximum then it will be unstable equilibrium, i.e,

$\frac{d^2U}{dx^2}<0$ .
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
Ans:

$(C),(D)$

A parallel plate capacitor of plate area A and plate separation d is charged to p.d. V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively the magnitude of charge on each plate, the electric field between the plates after insertion of dielectric and work done on the system in the process of inserting the slab, then-

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$Q=\dfrac {\epsilon_0 AV}{d}$
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$Q=\dfrac {\epsilon_0 KAV}{d}$
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$E=\dfrac {V}{Kd}$
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$W=\dfrac {\epsilon_0AV^2}{2d}(1-\dfrac {1}{K})$

Explanation

Before insert dielectric, capacitance,

$C_0=\dfrac{A\epsilon_0}{d}$
Magnitude of charge on each plate,

$Q=C_0V=\dfrac{A\epsilon_0}{d}V$
Before disconnect the battery, filed between the plates,

$E_0=\dfrac{V}{d}$
Afe insert dielectric, electric filed,

$E=\dfrac{E_0}{K}=\dfrac{V}{Kd}$
Before insert dielectric, work done,

$=W_i=\dfrac{1}{2}C_0V^2$
After insert work done,

$W_f=\dfrac{1}{2}(KC_0)(\dfrac{V}{K})^2$ .

$(C_f=\dfrac{A\epsilon_0K}{d}, V_f=\dfrac{E}{d})$
So net work done,

$W=W_i-W_f=\dfrac{1}{2}C_0V^2-\dfrac{1}{2}(KC_0)(\dfrac{V}{K})^2=\dfrac{1}{2}C_0V^2(1-\dfrac{1}{K})=\dfrac{A\epsilon_0V^2}{d}(1-\dfrac{1}{K})$

A parallel plate capacitor of plate area

$A$ and plate separation

$d$ is charged to potential difference

$V$ and then the battery is disconnected. A slab of dielectric constant

$k$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If

$Q,\space E$ and

$W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then:

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$Q = \displaystyle\frac{\epsilon_0 A V}{d}$
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$Q = \displaystyle\frac{\epsilon_0 k A V}{d}$
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$E = \displaystyle\frac{V}{k d}$
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$W = -\displaystyle\frac{\epsilon_0 A V^2}{2d}\left(1 - \displaystyle\frac{1}{k}\right)$

Explanation

As battery is removed, charge remains constant.

$Q_0=C_0V$

here

$C_0=\displaystyle\frac{A\epsilon_0}{d}$ ,

$V=\displaystyle\frac{Q_0d}{A\epsilon_0}$

After inserting dielectric, the field,

$E=\displaystyle\frac{Q_0}{Ak\epsilon_0}=\frac{V}{kd}$
and charge

$Q=Q_0=C_0V=\displaystyle\frac{A\epsilon_0V}{d}$ and

$C=kC_0$

The work done ,

$W=$ energy loss in capacitor

$=\displaystyle\frac{Q_0^2}{2C_0}-\frac{Q_0^2}{2kC_0}$

$W=\displaystyle\frac{Q_0^2}{2C_0}(1-\displaystyle\frac{1}{k})=1/2 C_0V^2(1-\displaystyle\frac{1}{k})=\displaystyle\frac{A\epsilon_0V^2}{2d}(1-\displaystyle\frac{1}{k})$

Note: sign convention , work done by the system is positive sign and work done on the system is negative sign.

A and B are two points in a closed circuit. The potential difference across the condenser of capacity

$5 \mu F$ is

0%
6 V
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10 V
0%
16 V
0%
4 V

Figure (i) shows a capacitors of capacitance

$C_1$ . A dielectric of dielectric constant K filling half the volume between the plates are inserted as shown in (ii) and (iii) having capacitances

$C_2$ and

$C_3$ . Then,

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$C_1=C_2=C_3$
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$C_2<C_1\ and\ C_3>C_1$
0%
$C_2>C_1 \space and \space C_3>C_1$
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$C_2>C_1 \space and \space C_3< C_1$

Explanation

For (i),

$C_1=\dfrac{A\epsilon_0}{d}$
For (ii), here two capacitors are in series- one is air filled capacitor with separation

$d/2$ and other is dielectric filled with separation

$d/2$ . Thus capacitance one is

$2C_1$ and other is

$2kC_1$ where k is dielectric constant.

$C_2=\dfrac{2C_1.2kC_1}{2C_1+2kC_1}=\dfrac{2kC_1}{k+1}$
For (iii), here two capacitors are in parallel- one is one is air filled capacitor with area

$A/2$ and other is dielectric filled with area

$A/2$ .Thus capacitance one is

$C_1/2$ and other is

$kC_1/2$

$C_3=C_1/2+kC_1/2=\dfrac{k+1}{2}C_1$
As

$k>1, C_2>C_1$ and

$C_3>C_1$

A parallel-plate air capacitor of capacitance

$C_0$ is connected to a cell of emf

$V$ and then disconnected from it. A dielectric constant

$K$ , which can just fill the air gap of capacitor, is now inserted in it. Which of the following is incorrect?

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The potential difference between the plates decreases $K$ times.
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The energy stored in the capacitor decreases $K$ times.
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The change in energy is $\dfrac{1}{2}C_0\varepsilon ^2 (K-1)$ .
0%
The change in energy is $\dfrac{1}{2}C_0\varepsilon ^2 \left ( 1-\dfrac{1}{k} \right )$ .

Explanation

As the capacitor is disconnected from cell so charge will remain constant .

here

$Q=C_0\epsilon$

Before inserting dielectric potential difference between the plates is

$V_0=\dfrac{Qd}{A\epsilon_0}$

after inserting, potential

$V=\dfrac{Qd}{Ak\epsilon_0}=\dfrac{V_0}{k}$

before inserting dielectric the energy is

$U_0=\dfrac{Q^2}{2C_0}$

after inserting, the energy is

$U=\dfrac{Q^2}{2kC_0}=\dfrac{U_0}{k}$

Change in energy

$=U_0-U=\dfrac{Q^2}{2C_0}-\dfrac{Q^2}{2kC_0}=\dfrac{Q^2}{2C_0}(1-\dfrac{1}{k})=\dfrac{(C_0\epsilon)^2}{2C_0}(1-\dfrac{1}{k})=\dfrac{1}{2}C_0\epsilon^2(1-\dfrac{1}{k})$

A parallel-plate air condenser of plate area A and separation d is charged to potential V and then the battery is removed. Now a slab of dielectric constant k is introduced between the plates. If Q, E and W denote respectively the magnitude of charge on each plate, the electric field between the plates (after introduction of dielectric slab) and work done on the system in the process of introducing the slab, then

0%
$\displaystyle W=\frac{\varepsilon _0 A Vh^2}{2d}(1-1/k)$
0%
$\displaystyle Q=\frac{\varepsilon _0 K A V}{d}$
0%
$\displaystyle Q=\frac{\varepsilon _0 A V}{d}$
0%
$\displaystyle E=\frac{V}{kd}$

Explanation

As battery is removed so charge remains constant.

$Q_0=C_0V$
here

$C_0=\frac{A\epsilon_0}{d}$ ,

$V=\frac{Q_0d}{A\epsilon_0}$
after insert dielectric, the filed,

$E=\frac{Q_0}{Ak\epsilon_0}=\frac{V}{kd}$
and charge

$Q=Q_0=C_0V=\frac{A\epsilon_0V}{d}$ and

$C=kC_0$
The work done ,

$W=$ energy loss in capacitor

$=\frac{Q_0^2}{2C_0}-\frac{Q_0^2}{2kC_0}$

$W=\frac{Q_0^2}{2C_0}(1-\frac{1}{k})=1/2 C_0V^2(1-\frac{1}{k})=\frac{A\epsilon_0V^2}{2d}(1-\frac{1}{k})$
Ans :(C),(D)

A fully charged capacitors has a capacitance

$'C'$ . It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity

$s$ and mass

$m$ . If the temperature of the block is raised by

$\Delta T$ , the potential difference

$V$ across the capacitor is :

0%
$\displaystyle \sqrt {\frac{2mC\Delta T}{s} }$
0%
$\displaystyle {\frac{mC\Delta T}{s} }$
0%
$\displaystyle {\frac{ms\Delta T}{C} }$
0%
$\displaystyle \sqrt {\frac{2ms\Delta T}{C} }$

Explanation

By conservation of energy, the stored energy in capacitor will transfer into heat energy during discharging. Thus,

$\displaystyle \frac{1}{2}CV^2=ms\Delta T \Rightarrow V=\sqrt{\frac{2ms\Delta T}{C}}$

What is the value of capacitance?

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$22.2 \mu F$
0%
$2.22 \mu F$
0%
$222 \mu F$
0%
none of these

Explanation

Let the capacitance is

$C$ .
Energy stored,

$U=\dfrac{1}{2}CV^2$

$\displaystyle C=\dfrac{2U}{V^2}=\dfrac{2\times 0.0160}{12^2}=2.22 \times 10^{-4} F=222 \mu F$

Two identical capacitors are connected as shown in given figure, having a charge

$q_0$ . A dielectric slab is introduced between the plates of the capacitor (I) so as to fill the gap, keeping the battery remain connected. The charge on each capacitors now will be :

0%
$\displaystyle\frac{2q_0}{\left [ 1+(1/k) \right ]}$
0%
$\displaystyle\frac{q_0}{\left [ 1+(1/k) \right ]}$
0%
$\displaystyle\frac{2q_0}{(1+k)}$
0%
$\displaystyle\frac{2q_0}{(1-k)}$

Explanation

Without dielectric :

$C_{eq}=\dfrac{C_0C_0}{C_0+C_0}=\dfrac{C_0}{2}$ and

$Q_{eq}=\dfrac{C_0}{2}\times 2V_0=C_0V_0$
As they are in series so the charge on each capacitors is equal to equivalent charge.i.e

$Q_{eq}=q_0=C_0V_0$
With dielectric in (I), the capacitance of it becomes

$kC_0$
now

$\displaystyle C'_{eq}=\dfrac{kC_0C_0}{kC_0+C_0}=\dfrac{k}{k+1}C_0$ and

$\displaystyle Q'_{eq}=\dfrac{k}{k+1}C_0(2V_0)=\dfrac{2k}{k+1}q_0$
Thus the on each capacitor is

$\displaystyle q'=Q'_{eq}=\dfrac{2k}{k+1}q_0=\dfrac{2q_0}{1+1/k}$

For the configuration of media of permitivities

${ \varepsilon}_{o },\ \varepsilon$ and

$\varepsilon_o$ between parallel plates each of area

$A$ , as show in figure, the equivalent capacitance is :

0%
${ \varepsilon }_{ 0 } A/d$
0%
${ \varepsilon }_{ 0 }\varepsilon A/d$
0%
$\cfrac { { \varepsilon }_{ 0 }\varepsilon A }{ d(\varepsilon +{ \varepsilon }_{ 0 }) }$
0%
$\cfrac { { \varepsilon }_{ 0 }\varepsilon A }{ (2\varepsilon +{ \varepsilon }_{ 0 }) }$

Explanation

Let the capacitance of the three capacitors be

${ C }_{ 1 },{ C }_{ 2 } \ \& \ { C }_{ 3 }$ respectively,

${ C }_{ 1 }=\cfrac { { \epsilon }_{ 0 }A }{ d } ,{ C }_{ 2 }=\cfrac { { \epsilon }A }{ d } \ \& \ { C }_{ 3 }=\cfrac { { \epsilon }_{ 0 }A }{ d }$
Now, they are connected in series, So the equivalent capacitance is

$\cfrac { 1 }{ C } =\cfrac { 1 }{ { C }_{ 1 } } +\cfrac { 1 }{ { C }_{ 2 } } +\cfrac { 1 }{ { C }_{ 3 } }$

$\Rightarrow \cfrac { 1 }{ C } =\cfrac { d }{ { \epsilon }_{ 0 }A } +\cfrac { d }{ \epsilon A } +\cfrac { d }{ { \epsilon }_{ 0 }A }$

$\cfrac { 1 }{ C } =\cfrac { d }{ A } \left[ \cfrac { 1 }{ \epsilon } +\cfrac { 2 }{ { \epsilon }_{ 0 } } \right]$

$\cfrac { 1 }{ C } =\cfrac { d\left( 2\epsilon +{ \epsilon }_{ 0 } \right) }{ A\epsilon { \epsilon }_{ 0 } }$

$\therefore C=\cfrac { { \epsilon }_{ 0 }\epsilon A }{ d\left( 2\epsilon +{ \epsilon }_{ 0 } \right) }$

An uncharged parallel plate capacitor having a dielectric of dielectric constant

$K$ is connected to a similar air cored parallel plate capacitor charged to a potential

${V}_{0}$ . The two share the charge, and the common potential becomes

$V$ . The dielectric constant

$K$ is :

0%
$\cfrac { { V }_{ 0 } }{ V } -1$
0%
$\cfrac { { V }_{ 0 } }{ V } +1$
0%
$\cfrac { V }{ { V }_{ 0 } } -1$
0%
$\cfrac { V }{ { V }_{ 0 } } +1$

A fully charged capacitor has a capacitance '

$C$ '. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity '

$s$ ' and mass '

$m$ '. If the temperature of the block is raised

$'\Delta T'$ , the potential difference '

$V$ ' across the capacitance is :

0%
$\dfrac {mC\Delta T}{s}$
0%
$\sqrt {\dfrac {2mC\Delta T}{s}}$
0%
$\sqrt {\dfrac {2ms\Delta T}{C}}$
0%
$\dfrac {ms\Delta T}{C}$

Explanation

Energy stored in the capacitor is

$U=\dfrac{1}{2}CV^2$ and
Heat energy in the wire is

$H=ms.\Delta T$
According to conservation of energy, the stored energy in the capacitor will be converted into heat energy.
So,

$U=H$

$\displaystyle \frac{1}{2}CV^2=ms.\Delta T$

$\displaystyle \therefore V=\sqrt{\dfrac{2ms.\Delta T}{C}}$

In the following diagram, the charge and potential difference across

$8 \mu F$ capacitance will be respectively (Approximately )

0%
$320 \mu C, 40 V$
0%
$420 \mu C, 50 V$
0%
$214 \mu C, 27 V$
0%
$360 \mu C, 45 V$

Two parallel plate capacitors of capacitances

$C$ and

$2C$ are connected in parallel and charged to a potential difference

$V$ . The battery is then disconnected, and the region between the plates of

$C$ is filled completely with a material of dielectric constant

$K$ . The common potential difference across the combination becomes :

0%
$\cfrac {2V}{K+2}$
0%
$\cfrac {V}{K+1}$
0%
$\cfrac {3V}{K+3}$
0%
$\cfrac {3V}{K+2}$

Explanation

Initially both capacitors are at potential

$V$ .
Charge on both plates are

$CV$ and

$2CV$ .
When we put dielectric between the plates of

$C$ ,

$C'=KC$
Common potential,

${ V }_{ C }=\cfrac { { Q }_{ 1 }+{ Q }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } =\cfrac { CV+2CV }{ KC+2C } =\cfrac { 3V }{ K+2 }$

The plates of a parallel plate capacitor are charged upto

$100V$ . now, after removing the battery, a

$2mm$ thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the capacitor plates is increased by

$1.6mm$ . The dielectric constant of the plate is :

0%
$5$
0%
$1.25$
0%
$4$
0%
$2.5$

Explanation

As battery is disconnected, so charge will remain the same.

It is given that final potential is the same.

So final capacitance should be

$C_1=C_2$

$\cfrac { { \varepsilon }_{ 0 }A }{ d } =\cfrac { { \varepsilon }_{ 0 }A }{ (1.6+d)-t(1-1/K) } \Rightarrow K = 5$

A dielectric slab fills the lower half of a parallel plate capacitor as shown in figure :

(Take plate are as

$A$ )

0%
equivalent capacity of the system is $\big (({ \varepsilon }_{ 0 }A/2d)(1+K)\big )$
0%
the net charge of the lower half of the left hand plate is $1/K$ times the charge on the upper half of the plate
0%
net charges on the lower and upper halves of the left hand plate are different
0%
net charge on the lower hand of the left hand plate is $\cfrac { K{ \varepsilon }_{ 0 }A }{ 2d } \times V$

Explanation

This system can be considered as two capacitors in parallel with

${C}_{1}=\cfrac { { \varepsilon }_{ 0 }A }{ 2d }$ and

${C}_{2}=\cfrac { { K\varepsilon }_{ 0 }A }{ 2d }$

$C_{eq}=C_1+C_2=\cfrac { { \varepsilon }_{ 0 }A }{ 2d }+\cfrac { { K\varepsilon }_{ 0 }A }{ 2d }=\cfrac{\epsilon_0 A}{2d}\times(1+K)$ . Hence option A is correct

${C}_{1}$ is due to upper half of two plates, while

${C}_{2}$ is due to lower half. As potential difference across two capacitors is the same, charges would be different as capacitors are different. Hence option C is correct,

Net charge on the lower half plate

$Q_2=C_2V=\cfrac { { K\varepsilon }_{ 0 }A }{ 2d }V$ .

Therefore option D is correct

An electric charge

$10^{-3}\mu C$ is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at

$(\sqrt 2, \sqrt 2)$ and (2, 0) respectively. The potential difference between the points A and B will be :

0%
4.5 volt
0%
9 volt
0%
zero
0%
2 volt

Explanation

The distance of point A from origin is

$r_A=\sqrt{(\sqrt 2-0)^2+(\sqrt 2-0)^2}=\sqrt{4}=2$
The distance of point B from origin is

$r_B=\sqrt{(2-0)^2+(0-0)^2}=\sqrt{4}=2$
Potential at A due to charge

$Q=10^{-3} \mu C$ is

$\displaystyle V_A=\frac{Q}{4\pi\epsilon_0 r_A}=\frac{Q}{8\pi\epsilon_0 }$ as

$r_A=2$
and potential at B is

$\displaystyle V_B=\frac{Q}{4\pi\epsilon_0 r_B}=\frac{Q}{8\pi\epsilon_0 }$ as

$r_B=2$
Potential difference between A and B

$\displaystyle=V_A-V_B=\frac{Q}{8\pi\epsilon_0 }-\frac{Q}{8\pi\epsilon_0 }=0$ volt.

The electric field in the dielectric slab is:

0%
$\dfrac {V}{Kd}$
0%
$\dfrac {KV}{d}$
0%
$\dfrac {V}{d}$
0%
$\dfrac {KV}{t}$

Explanation

Initially, the electric filed is

$E_0=\dfrac{Q}{A\epsilon_0}$

also,

$E_0=\dfrac{V}{d}$

As battery is disconnected so charge Q is constant.

After inserted the dielectric the electric field becomes

$E'=\dfrac{Q}{AK\epsilon_0}=\dfrac{E_0}{K}=\dfrac{V}{Kd}$

An air capacitor of capacity

$C=10\mu F$ is connected to a constant voltage battery of

$12$ volt. Now the space between the plates is filled with a liquid of dielectric constant

$5$ . The (additional) charge that flows now from battery to the capacitor is :

0%
$120 \mu C$
0%
$600 \mu C$
0%
$480 \mu C$
0%
$24 \mu C$

Explanation

Initial charge

$q_i=CV=10(12)=120 \mu C$ .
After filled with dielectric liquid the charge

$q'=C'V=KCV=5(10)12=600 \mu C$
The (additional) charge that flows now from battery to the capacitor

$=q_f-q_i=600-120=480 \mu C$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

By inserting dielectric slab, value of

$C_2$ will increase to

$KC_2$ where K is the dielectric constant. In series potential difference distributes in inverse ratio of capacitance. As capacitance

$C_2$ is increased so potential across

$C_2$ will decrease.
If

$C_2$ is increased, charge on capacitors will also increase.

$(q=CV)$
So, positive charge or current flows in clockwise direction.

Figure shows a parallel plate capacitor with plate area

$A$ and plate separation

$d$ . A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dielectric constant

$K$ is placed in between the plates of the capacitor as shown.
Now, answer the following questions based on above information. The electric field in the gaps between the plates and the dielectric slab will is :

0%
$\dfrac {\varepsilon AV}{d}$
0%
$\dfrac {V}{d}$
0%
$\dfrac {KV}{d}$
0%
$\dfrac {V}{d-t}$

Explanation

The gap between the plates and the dielectric field is filled with air/vacuum. So electric field here is same as electric field with vacuum between the parallel plate capacitor, i,e,

$E_o=V/d$

Two parallel plate capacitors of capacitances

$C$ and

$2C$ are connected in parallel and charged to a potential difference

$V$ . The battery is then disconnected and the region between the plates of the capacitor

$C$ is completely filled with a material of dielectric constant

$K$ . The potential difference across the capacitors now becomes :

0%
$\dfrac {3V}{K+2}$
0%
$KV$
0%
$\dfrac {V}{K}$
0%
$\dfrac {3}{KV}$

Explanation

Initial total charge of the system is:

$Q_i=Q_1+Q_2=CV+2CV=3CV$
When dielectric is inserted in

$C$ so the capacitance becomes

$KC$
Final charge,

$Q_f=Q'_1+Q'_2=KC V'+2CV'=(K+2)CV'$ where

$V'=$ common potential after disconnected the battery.
As battery is disconnected so total charge will remain unchanged.
Thus,

$Q_i=Q_f \Rightarrow 3CV=(K+2)CV'$

$\therefore V'=\dfrac{3V}{K+2}$

A parallel-plate air capacitor of capacitance

$245 pF$ has a charge of magnitude

$0.148\mu C$ on each plate. The plates are

$0.328 mm$ apart. What is the surface charge density on each plate?

0%
$8.7\mu C/m^2$
0%
$16.3\mu C/m^2$
0%
$8.7 mC/m^2$
0%
$16.3 pC/m^2$

Explanation

The capacitance of parallel plate air capacitor is

$C=\dfrac{A\epsilon_0}{d}$

So, area of each plate

$A=\dfrac{Cd}{\epsilon_0}=\dfrac{(245\times 10^{-12})(0.328\times 10^{-3})}{8.85\times 10^{-12}}=9.08 \times 10^{-3}m^2$

Thus, surface charge density,

$\sigma=\dfrac{Q}{A}=\dfrac{0.148\times 10^{-6}}{9.08\times 10^{-3}}=16.3\mu C/m^2$

$40 \mu F$ capacitor in defibrillator is charged to 3000 V. the energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. the power delivered to patient is :

0%
45 kW
0%
180 kW
0%
90 kW
0%
360 kW

Two unlike charges of magnitude q are separated by a distance 2d. The potential at a point midway between them is

0%
zero
0%
$\dfrac{1}{4 \pi {\epsilon}_{0}}$
0%
$\dfrac{1}{4 \pi {\epsilon}_{0}}$ . $\dfrac{q}{d}$
0%
$\dfrac{1}{4 \pi {\epsilon}_{0}}$ . $\dfrac{2q}{d}$

A parallel plate capacitor of plate area

$A$ and plate separation

$d$ is charged to potential difference

$V$ and then the battery is disconnected. A slab of dielectric constant

$K$ is then inserted between the plates of capacitor so as to fill the space between the plates. If

$Q, E$ and

$W$ denote respectively, the magnitude of charge on each plate electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then which is wrong?

0%
$Q=\dfrac {\epsilon_0AV}{d}$
0%
$Q=\dfrac {\epsilon_0KAV}{d}$
0%
$E=\dfrac {V}{Kd}$
0%
$W=\dfrac {\epsilon_0AV^2}{2d}\left (1-\dfrac {1}{K}\right )$

Explanation

Initial capacitance,

$\displaystyle C_0=\dfrac{A\epsilon_0}{d}$
So initial charge on the capacitor is:

$Q_0=C_0V=\dfrac{AV\epsilon_0}{d}$
As the battery is disconnected so charge remains unchanged.

So,

$\displaystyle Q=Q_0=\dfrac{AV\epsilon_0}{d}$
After insert the dielectric the capacitance becomes,

$C'=KC_0$ where K dielectric constant.
Now potential becomes,

$V'=Q/C'=Q_0/KC_0=V/K$
Field,

$E=V'/d=\dfrac{V}{Kd}$
Initial energy,

$\displaystyle U_0=\frac{1}{2}C_0V^2=\frac{AV^2\epsilon_0}{2d}$ and
final energy

$\displaystyle U_f=\frac{1}{2}C'V'^2=(1/2)(KC_0)(V/K)^2=\frac{AV^2\epsilon_0}{2Kd}$

Since, work done

$=$ Decrease in Potential Energy
Work done,

$\displaystyle W=-\Delta U=-(U_f-U_0)=U_0-U_f=\frac{1}{2} \frac{AV^2\epsilon_0}{d}(1-1/K)$

A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by

$Q_0, V_0, E_0$ , and

$U_0$ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by

$Q, V, E$ and

$U$ are related to the previous ones are :

0%
$Q > Q_0$
0%
$V > V_0$
0%
$E > E_0$
0%
$U < U_0$

Explanation

As the battery till in connection so potential will remained unchanged. i,e

$V=V_0$

$E_0=\dfrac{V_0}{d}$ and

$E=\dfrac{V}{d}=\dfrac{V_0}{d} \Rightarrow E=E_0$ where

$d=$ separation of plates.
Let initial capacitance

$=C_0$
When dielectric slab with dielectric constant K is inserted, the capacitance becomes

$C=KC_0$ .
Thus,

$Q_0=C_0V_0$ and

$Q=CV=KC_0V_0=KQ_0 \Rightarrow Q>Q_0$ as

$(K>1)$
Also,

$U_0=(1/2)C_0V_0^2$ and

$U=(1/2)CV^2=(1/2)KC_0.V_0=KU_0$
Hence,

$U>U_0$

Identical charges -q each are placed at 8 corners of a cube of each side b. Electrostatic potential energy of a charge +q which is placed at the centre of cube will be :

0%
$\dfrac {-4\sqrt 2q^2}{\pi \epsilon_0b}$
0%
$\dfrac {-8\sqrt 3q^2}{\pi \epsilon_0b}$
0%
$\dfrac {-4q^2}{\sqrt 3\pi \epsilon_0b}$
0%
$\dfrac {-8\sqrt 2q^2}{\pi \epsilon_0b}$

Explanation

Here,

$AD=CD=OC=b/2$ and

$AC^2=AD^2+CD^2=\dfrac{b^2}{4}+\dfrac{b^2}{4}=\dfrac{b^2}{2}$

Thus distance of each charge from centre is

$OA=r=\sqrt{AC^2+OC^2}=\sqrt{\dfrac{b^2}{2}+\dfrac{b^2}{4}}=\dfrac{\sqrt{3}b}{2}$

Potential energy at the centre is

$\displaystyle W=\dfrac{8(-q)q}{4\pi\epsilon_0 r}=\dfrac{8(-q)q}{4\pi\epsilon_0 (\frac{\sqrt{3}b}{2})}=\frac{-4q^2}{\sqrt{3}\pi\epsilon_0 b}$

Which capacitor has the largest potential difference between its plates ?

0%
A
0%
B
0%
D
0%
A and D are the same and larger than B or C

Explanation

For A: the capacitance

$C_A=\dfrac{A\epsilon_0}{d}$
For B: the capacitance

$C_B=\dfrac{A\epsilon_0}{2d}$
For C: the capacitance

$C_C=\dfrac{Ak\epsilon_0}{d}=\dfrac{5A\epsilon_0}{d}$
For D: the capacitance

$C_D=\dfrac{Ak\epsilon_0}{2d}=\dfrac{5A\epsilon_0}{2d}$
so,

$C_B<C_A<C_D<C_C$
Potential difference ,

$V=\dfrac{Q}{C}$

So, the capacitor with smallest capacitance will have largest potential difference between the plates.

Here B has smallest capacitance, so B will have largest potential difference.

The following operation can be performed on a capacitor:
X connect the capacitor to a battery of emf E.
Y disconnect the battery.
Z reconnect the battery with the polarity reversed.
W insert a dielectric slab in the capacitor.

0%
In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed
0%
The charge appearing on the capacitor is greater after the action XWY than after the action XYW.
0%
The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
0%
The electric field in the capacitor after the action XW is the same as that after WX.

Explanation

For option (B) solution : In XWY, charge on capacitor plate =KCE

In XYW, charge on capacitor plate = CE

For option (C) solution : In WXY, U = $\displaystyle \frac {1} {2} KCE^2$

In XYW, U = $\displaystyle \frac {CE^2} {2K}$

For option (D) solution : 2 both cases, electric field = $\displaystyle \frac {E} {d}$ .

Positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point A at distance r(r> R) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R/2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion.

0%
$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$
0%
$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{8}\right) \right]^{1/2}$
0%
$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} - \frac{3}{8}\right) \right]^{1/2}$
0%
$\displaystyle \left[\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$

Explanation

Initial energy are kinetic and potential energy given by-

$K_i=\dfrac{1}{2}mv^2$ and

$U_i=\dfrac{Qq}{4\pi\epsilon_o r}$

Finally at last point , its velocity get reduced to

$0$ , and potential at a point inside sphere is given by-

$V=\dfrac{Q}{4\pi\epsilon_o}\dfrac{3R^2-r^2}{2R^3}$

$r=\dfrac{R}{2}$ ,

$U_f=qV$

$\implies U_f=\dfrac{Qq}{4\pi\epsilon_o}\dfrac{3R^2-\dfrac{R^2}{4}}{2R^3}$

$\implies U_f=\dfrac{11}{8}\dfrac{Qq}{4\pi\epsilon_oR}$

Now applying conservation of mechanical energy-

$K_i+U_i=K_f+U_f$

$\implies \dfrac{1}{2}mv^2+\dfrac{Qq}{4\pi\epsilon_o r}=0+\dfrac{11}{8}\dfrac{Qq}{4\pi\epsilon_o R}$

$\implies mv^2=\dfrac{Qq}{2\pi\epsilon_o}\left(\dfrac{11}{8R}-\dfrac{1}{r}\right)$

$\implies v^2=\dfrac{Qq}{2\pi\epsilon_oRm}\left(\dfrac{11}{8}-\dfrac{R}{r}\right)$

$\implies v^2=\dfrac{Qq}{2\pi\epsilon_oRm}\left(1+\dfrac{3}{8}-\dfrac{R}{r}\right)$

$\implies v^2=\dfrac{1}{2\pi\epsilon_o}\dfrac{Qq}{Rm}\left(\dfrac{r-R}{r}+\dfrac{3}{8}\right)$

$\implies v=\sqrt{\dfrac{1}{2\pi\epsilon_o}\dfrac{Qq}{Rm}\left(\dfrac{r-R}{r}+\dfrac{3}{8}\right)}$

Answer-(B)

Which list below places the capacitors in order of increasing capacitance ?

0%
A, B, C, D
0%
B, A, C, D
0%
B, A, D, C
0%
A, B, D, C

Explanation

For A: the capacitance

$C_A=\dfrac{A\epsilon_0}{d}$

For B: the capacitance

$C_B=\dfrac{A\epsilon_0}{2d}$

For C: the capacitance

$C_C=\dfrac{Ak\epsilon_0}{d}=\dfrac{5A\epsilon_0}{d}$

For D: the capacitance

$C_D=\dfrac{Ak\epsilon_0}{2d}=\dfrac{5A\epsilon_0}{2d}$

so,

$C_B<C_A<C_D<C_C$

A parallel plate capacitor of area A and separation

$d$ is charged to potential difference

$V$ and removed from the charging source A dielectric slab of constant

$K = 5$ , thickness

$d$ and area

$\displaystyle \frac{A}{3}$ is inserted as shown in the figure Let

$\displaystyle \sigma _{1}$ be free charge density at the conductor-dielectric surface and

$\displaystyle \sigma _{2}$ be the charge density at the conductor-vacuum surface :

0%
the electric field have the same value inside the dielectric as in the free space between the plates
0%
the ratio $\displaystyle \frac{\sigma _{1}}{\sigma _{2}}$ is equal to $\displaystyle \frac{1}{5}$
0%
the new capacitance is $\displaystyle \frac{7\varepsilon _{0}A}{3d}$
0%
the new potential difference is $\displaystyle \frac{3}{7}V$

Explanation

Potential for each plate remain same over whole are if potential difference thems is way V then V = Ed i.e. E is also same inside the plates

To Keep E same free change dansity is changed i.e. charge redistributes itself

To find new capacitance, two capacitors can be taken as connected in parallel Then

$\displaystyle C_{eq}=\frac{5.\epsilon _{0}A/3}{d}+\frac{\epsilon _{0}.2A/3}{d}=\frac{7\epsilon _{0}A}{3d}$

Q = CV, as Q remains unchanged V is changed to $\displaystyle \frac{3}{7}V$

Choose the incorrect statement:-

0%
the potential energy per unit positive charge in an electric field at some point is called the electric potential
0%
the work required to be done to move a point charge from one point to another in an electric field depends on the position of the points.
0%
the potenial energy of the system will increase if a positive charge is moved against of Coulombian force
0%
the value of funcdamental charge is not equivalent to the electronic charge

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 12 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 12 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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