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Parallel plate capacitor - class-XII

Description: parallel plate capacitor
Number of Questions: 30
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Tags: electrostatic potential and capacitance physics electrostatics
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Choose the correct answer from the alternatives given.
The charge on a parallel plate capacitor varies as $q \, = \, q _0 \, cos2\pi \nu t$. The plates are very large and close together (area = A, separation = d). The displacement current through the capacitor is then

  1. $3q _0 \, 2\pi \nu \, sin2\pi \nu t$

  2. $-q _0 \, 2\pi \nu \, sin2\pi \nu t$

  3. $4q _0 \, 2\pi \, sin2\pi \nu t$

  4. $5q _0 \, 2\pi \, sin2\pi \nu t$


Correct Option: B
Explanation:

Given: The charge on the parallel plate capacitor varies as $q=q _0cos2\pi\nu t$


To find: The displacement current through the capacitor.

The displacement current in a capacitor is equal to the conduction current of the capacitor.
Displacement current, $I _D \, =I _C$


The conduction current in a capacitor is given by:
$I _C= \dfrac{dq}{dt}$
$\implies\, \dfrac{d}{dt} \, [q _0 \, cos \, 2\pi\nu t]$

$  \,=\, -q _0 \, 2\pi \nu\, sin2\pi \nu\,t$


Option $(B)$ is correct.

The capacitance of a capacitor does not depend on

  1. the separation between the plates

  2. the size of the plates

  3. the charges on the plates

  4. none of the above


Correct Option: D
Explanation:

$C=\cfrac{Q}{V}=\cfrac{\epsilon _oA}{d}$

Capacitance depends on charge, size and separation between the plates.

Three connected conductors A, B and C have a total charge of 48$\mu V$. The ratio of their capacitance are 1 :3 : 2. The charges on thei individually.

  1. 24$
    \mu C
    $
    , 12$
    \mu C
    $
    , 12$
    \mu C
    $

  2. 8$
    \mu C
    $
    ,18$
    \mu C
    $
    ,22$
    \mu C
    $

  3. 8$
    \mu C
    $
    ,24$
    \mu C
    $
    ,16$
    \mu C
    $

  4. 16$
    \mu C
    $
    , 16$
    \mu C
    $
    ,16$
    \mu C
    $


Correct Option: C
Explanation:

Three conductors $A , B , C$ have total charge $Q=48 \ \mu C$

Let charge on individual capacitors be $Q _1 , Q _2 , Q _3$
$\therefore Q _1+Q _2+Q _3=Q=48 \ \mu C ... 1$
also $C _1:C _2:C _3::1:3:2 ... 2$
For three conductors are in parallel $\dfrac{Q _1}{C _1}=\dfrac{Q _2}{C _2}=\dfrac{Q _3}{C _3} ... 3$
From $2 \  and \ 3$ we get
$3Q _1=Q _2 \ and \ 2Q _1=Q _3 ... 4$
substituting $4 \  in \ 1$
$\therefore Q _1+3Q _1+2Q _1= 48 \ \mu C$
$\therefore Q _1=8 \ \mu C$ 
substituting in $4$ we get 
$Q _2=24 \ \mu C \ and \ Q _3=16 \ \mu C$
Hence the charge on individual capacitors are $Q _1=8 \ \mu C ,Q _2=16 \ \mu C , Q _3=24 \ \mu C$

A capacitor of $40\ \mu F$ charged upto $1000\ V$ is joined in parallel to another capacitor of $20\ \mu F$ charged upto $400\ V$. What is the common potential difference between the two ends of their connection ?

  1. $900\ V$

  2. $700\ V$

  3. $850\ V$

  4. $1000\ V$


Correct Option: C

Two capacitor each having a capacitance $C$ and breakdown voltage $V$ are joined in series. The effective capacitance and maximum working voltage of the combination is:-

  1. $2C, 2V$

  2. $\dfrac{C}{2}, \dfrac{V}{2}$

  3. $2C, V$

  4. $\dfrac{C}{2}, 2V$


Correct Option: D
Explanation:
In series arrangement at charge on each plate of each capacitor has the same magnitude the potential difference in distributed inversely in the ratio of capacitor ie,

V=V1+V2
V=2V

The equivalent capacitance C's is given by
1/Cs=1/C1+1/C2
Cs=C/2

answer is
2V and C/2

A cylindrical capacitor has two co-axial cylinders of length $20\ cm$ and radii $2r$ and $r$. Inner cylinder is given a charge $10\ \mu F$. The potential difference between the two cylinders will be ?

  1. $\dfrac {0.1 \ln {2}}{4 \pi \epsilon _{0}}m\ V$

  2. $\dfrac {\ln {2}}{4 \pi \epsilon _{0}}m\ V$

  3. $\dfrac {10\ln {2}}{4 \pi \epsilon _{0}}m\ V$

  4. $\dfrac {0.01\ln {2}}{4 \pi \epsilon _{0}}m\ V$


Correct Option: A

If on combining two charged bodies, the current does not flow then :

  1. charge is equal on both

  2. capacitance is equal on both

  3. potential is equal on both

  4. resistance is equal on both


Correct Option: C
Explanation:

The current means the charge will flow from one region to other region due to the potential difference between two regions. So if the potential is equal on both , the current will not flow.

A condenser of capacity $ 2 \mu F$ is charged to a potential of 200V. It is now connected to an uncharged condenser of capacity $ 3 \mu F$. The common potential is :

  1. 200 V

  2. 100 V

  3. 80 V

  4. 40 V


Correct Option: C
Explanation:
Potential difference is same
$V=\dfrac{q}{C}$
$\dfrac{q _1}{C _1} = \dfrac{q _2}{C _2}=V$
$\dfrac{q _1}{2}=\dfrac{q _2}{C}=V$

By conservation of charge, $q _1+q _2=400$

Solving the above two equations give:
$q _1 = 160 \mu C$
$q _2=240 \mu C$

$V=\dfrac{q _2}{C _2}=80 V$

Two connected bodies having respectively capacitances ${\text{C}} _{\text{1}} \,{\text{and}}\,{\text{C}} _{\text{2}} $ are charged with a total charge Q. The potentials of the two bodies are.

  1. $
    \dfrac{{\text{Q}}}
    {{{\text{C}} _{\text{1}} + C _2 }},\dfrac{Q}
    {{C _1 + C _2 }}
    $

  2. $
    \dfrac{Q}
    {{C _1 }} + \dfrac{{\text{Q}}}
    {{C _2 }},\dfrac{Q}
    {{C _1 }} + \dfrac{{\text{Q}}}
    {{C _3 }}
    $

  3. $
    \dfrac{{C _1 C _2 }}
    {{C _1 + C _2 }},\dfrac{{C _1 C _2 }}
    {{C _1 - C _2 }}
    $

  4. $
    \dfrac{Q}
    {{C _1 }} - \dfrac{Q}
    {{C _2 }},\dfrac{Q}
    {{C _1 - C _2 }}
    $


Correct Option: A
Explanation:
Given that
Charge on the body $=Q$
Now, the capacitance$=C _1$ and $C _2$
Again we know that
$V=\cfrac{Q}{C}$
$\therefore$ the total capacitance here $=C _1+C _2$
$V=\cfrac{Q}{C _1+C _2}$
$\therefore$ The potential for both $=(\cfrac{Q}{C _1+C _2})(\cfrac{Q}{C _1+C _2})$

Two capacitors A and B of capacitance $ 6 \mu F$ and $10 \mu F$ respectively are connected in parallel and this combination is connected in series with a third capacitors C of $ 4 \mu F $. A potential difference of 100 volt is applied across the entire combination. Find the charge and potential difference across $6\ \mu F$ capacitor.

  1. $120 \mu \, C; 20 V.$

  2. $200 \mu \, C; 20 V.$

  3. $320 \mu \, C; 80 V.$

  4. $320 \mu \, C; 60 V.$


Correct Option: A
Explanation:
Since A and B are connected and Parallel,
Hence,
Equivalence capacitance of capacitor$ A$ and $B = 6\mu F +10\mu F = 16\mu F$ 

Now $16\mu F$ and capacitor C of $4\mu F$ are connected in series
Hence,
the equivalence capacitance $(EC)$ will be given by
$\dfrac{1}{EC}= \dfrac {1}{16} +\dfrac {1}{4}$
we get,
Equivalence capacitance $(EC) = \dfrac{16}{5} \mu F$
$Charge = EC\times P.D.$
$Charge = \dfrac{16}{5}\times 100$
$Charge(Q) = 320\ \mu F $

Now,
$Q _A=320\times \dfrac {6}{16} \mu F$
$Q _A=120 \ \mu F$
$Q _B=320\times \dfrac {10}{16} \mu F$
$Q _B=200 \ \mu F$

Now proceeding for P.D. across each capacitor,
$V _A=\dfrac{Q _A}{C _A}$

$V _A=\dfrac{120}{6} = 20 V$ ,

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

$V _B=\dfrac{200}{10} = 20 V$ ,

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

$V _C=\dfrac{320}{4}=80 V$
this is the required solution.

A glass slab is put within the plates of a charged parallel plate condenser. Which of the following quantities does not change?

  1. energy of the condenser

  2. capacity

  3. intensity of electric field

  4. charge


Correct Option: B

Find the potential at a point due to a positive charge of $100\mu C$ at a distance of $10\ m$ in a medium of dielectric constant $9$.

  1. $10^{7}V$.

  2. $10^{4}V$.

  3. $10^{5}V$.

  4. $10^{6}V$.


Correct Option: B

A capacitor of capacitance 2 $\mu $ F is charged to a potential difference of 12 V. It is then connected across inductor of inductance 0.6 mH. The current in the circuit when the potential difference across the capacitor 6 V is :

  1. 3.6 A

  2. 2.4 A

  3. 1.2 A

  4. 0.06 A


Correct Option: D

The parallel plates of capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. Alter 1 second the potential difference between the plates of the capacitor is 240 volts then after 2 seconds the potential difference between the plates will be -

  1. 200 V

  2. 180 V

  3. 160 V

  4. 140 V


Correct Option: B

When charge is supplied in a system where the capacitor are connnected in parallel. the charge will be distributed

  1. Equally to all the capacitor

  2. indirect ratio to the capacitance value

  3. in an inverse ratio to the capacitance values

  4. both (b) and (c)


Correct Option: C

Two identical capacitors are connected in parallel across a potenial difference V. after they are fully charged, the positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to positive plate of other. The loss of energy will be

  1. $\dfrac { 1 }{ 2 } { CV }^{ 2 }$

  2. ${ CV }^{ 2 }$

  3. $\dfrac { 1 }{ 4 } { CV }^{ 2 }$

  4. Zero


Correct Option: C

A simple pendulum of mass m charged negatively to q coulomb oscillates with a time period T in a downward electric field E such that mg > qE. If the electric field is withdrawn, the new time period :

  1. $=$T

  2. $>$T

  3. $<$T

  4. any of the above three is possible


Correct Option: C
Explanation:

Time period in the absence of electric field $T' = 2 \pi \displaystyle \sqrt{\frac{l}{g}}$

In the presence of electric field
$g _{eff}=(mg - qE) / m$
Therefore $T = 2 \pi \displaystyle \sqrt{\frac{l}{(mg - qE) / m}}$
$\displaystyle T'  = 2\pi \sqrt{\frac{l}{g}}$
or $T' < T$

Among two discs $A$ and $B$, first have radius $10\ cm$ and charge ${10}^{-6}\ \mu C$ and second have radius $30\ cm$ and charge ${10}^{-5}C$. When they are touched, charge on both ${q} _{A}$ and ${q} _{B}$ respectively will be :

  1. ${q} _{A}=2.75\mu C,{q} _{B}=3.15\mu C$

  2. ${q} _{A}=1.09\mu C,{q} _{B}=1.53\mu C$

  3. ${q} _{A}={q} _{B}=5.5\mu C$

  4. none of these


Correct Option: C

Two metal pieces having a potential difference of 800 V are 0.02 m apart horizontally. A particle of mass $1.96\times 10^{-15}kg$ is suspended in equilibrium between the plates. If e is the elementary charge, then charge on the particle is

  1. 8

  2. 6

  3. 0.1

  4. 3


Correct Option: D
Explanation:

Force due to gravity on the particle is $F _g=mg$
Force due to field E between the metal plates is $\displaystyle F _e=qE=(ne)\frac{V}{d}$.
In equilibrium, $\displaystyle F _g=F _e \Rightarrow mg=(ne)\frac{V}{d}$


$\displaystyle \therefore n=\dfrac{mgd}{eV}$

$=\dfrac{1.96\times 10^{-15}\times 9.8\times 0.02}{1.6\times 10^{-19}\times 800}=3$

Consider two bodies A and B of same capacitance.  If charge of -10C flows from body A to body B, then

  1. the potential of body A increases.

  2. the potential of body B decreases

  3. the magnitude of change in potential in both bodies is same.

  4. All the above


Correct Option: D
Explanation:

As we know $ V = \dfrac{Q}{C}$ , where letters have their respective meanings.

So, if charge of $-10C$  (Negative Charge) flows from body A to body B, then the potential of body A increases and the potential of body B decreases. Also the magnitude of change in potential in both bodies is same as A and B are of same capacitance.
Therefore, D is correct option.

Which of the following is true about field between parallel charged plates?

  1. It is strongest between the plates

  2. It is strongest near the positive plate

  3. It is strongest near the negative plate

  4. The field is constant between the plates

  5. The field is variable, therefore the strong point also varies


Correct Option: D
Explanation:

The field is constant between the plates , because we dealing with a system of two charged plates.we know the electric field between two charged plates (suppose both have positive charge) is given by 

                   $E=\frac{1}{2\varepsilon _{0}}\left(\sigma _{1}-\sigma _{2}\right)$         where $\sigma=$ surface charge density
we can see that electric field doesn't depend upon distance from plates , therefore it is constant.

A capacitor contains two square plates with side lengths $5.0$ cm. The plates are separated by $2.0$ mm. Dry air fills the space between the plates. Dry air has a dielectric constant of $1.00$ and experiences dielectric breakdown when the electric field exceeds $3.0 \times  10^4$ V/cm.
What is the magnitude of charge that can be stored on each plate before the capacitor exceeds its breakdown limit and sends a spark between the plates?

  1. $6.6 \times  10^{-8}C$

  2. $6.6 \times 10^{-5}C$

  3. $3.3 \times  10^{-7}C$

  4. $3.3 \times  10^{-84}C$

  5. $8.1 \times  10^{-2}C$


Correct Option: A
Explanation:

Given :   $d = 2$  mm             $E = 3.0 \times 10^4$ $V/cm = 3.0 \times 10^6$  $V/m$         $l = 5.0 cm  = 0.05$ m

Area of each plate       $A = l^2 = (0.05)^2  = 25 \times 10^{-4}$  $m^2$
Capacitance of the capacitor         $C = \dfrac{A\epsilon _o }{d}$

Potential difference between the plates         $V = Ed$
$\therefore$   Charge on each plate      $Q = CV = A\epsilon _o E$
$\implies$   $Q = 25\times 10^{-4} \times 8.85 \times 10^{-12} \times 3.0 \times 10^6  = 6.6 \times 10^{-8}$  C

An air-gap parallel plate capacitor is fully charged by a battery.
What combination of two measurements will allow someone to calculate the magnitude of the electric field in between the capacitor plates?

  1. The potential difference of the battery and the area of the plates.

  2. The charge on the plates and the distance between the plates.

  3. The charge on the plates and the area of the plates.

  4. The area of the plates and the distance between the plates.

  5. More than two measurements are needed to calculate the electric field in between the capacitor plates.


Correct Option: C
Explanation:

Capacitance of the parallel plate capacitor       $C = \dfrac{A\epsilon _o}{d}$

Using     $Q = CV$               $\implies V = \dfrac{Q}{C}$
 Electric field between the plates      $E = \dfrac{V}{d}$
$\therefore$   $E =\dfrac{Q}{C d} = \dfrac{Q}{\frac{A\epsilon _o}{d} \times d} = \dfrac{Q}{A\epsilon _o}$
Thus option C is correct.

Two capacitors of $10\ pF$ and $20\ pF$ are connected to $200\ V$ and $100\ V$ sources respectively. If they are connected in parallel by the wire, what is the common potential of the capacitors?

  1. $133.3\ Volt$

  2. $150\ Volt$

  3. $300\ Volt$

  4. $400\ Volt$


Correct Option: A
Explanation:

Total charge on the two capacitors is:

$Q = Q _1 +Q _2 = C _1V _1+C _2V _2 $
$Q = 10 \times 10^{-9} \times 200 + 20 \times 10^{-9} \times 100$
$Q = 4\ \mu C$

Net capacitance of two capacitors in parallel is:
$C = C _1+C _2$
$C = 30\ pF$

Common potential of the parallel combination of capacitors is:
$V = \cfrac{Q}{C}$
$V = \cfrac{4\times 10^{-6}}{30 \times 10^{-9}} = 133.3\ V$

A thunder cloud and the earth's surface may be regarded as a pair of charged parallel plates separated by a distance $h$ and the capacitance of the system is $C$. When a flash of mean current '$i$' occurs for a time duration '$t$', the electric field strength between the cloud and earth is:

  1. $\dfrac { it }{ C } $

  2. $Cit$

  3. $\dfrac { it }{ Ch } $

  4. $\dfrac { Cit }{ h } $


Correct Option: C
Explanation:

Total charge accumulation in time duration $t$ is:

$Q = it$

From definition of capacitance, voltage induced due to this charge is:
$V = \cfrac{Q}{C}$
$V = \cfrac{it}{C}$

Electric field is defined as negative of gradient of potential. Hence,
$E = -\cfrac{dV}{dr}$
$\left| E \right| = \cfrac{V}{h}$
$\left| E \right| = \cfrac{it}{Ch}$

You measure the capacitor and inductor voltages in a driven RLC circuit, and find 10V for the rms capacitor voltage and 15V for the rms inductor voltage.

  1. $\omega = \omega _{res}$

  2. $\omega < \omega _{res}$

  3. $\omega > \omega _{res}$

  4. Can't be said


Correct Option: C
Explanation:
A/c to ques $V _{c}=10\ V$ and $V _{2}=15\ V$ 
Let frequency be $W$ capacitance be $C$ and inductance be $L$
$\Rightarrow iX _{C}=10$ and $iX _{L}=15$
$\Rightarrow \dfrac{i}{WL}=10$ and $i(WL)=15$
Dividing both
$\dfrac{\dfrac{i}{WC}}{i VWL}=\dfrac{10}{15}$
$\Rightarrow W^{2}=\dfrac{3/2}{LC}$
$W=\dfrac{\sqrt{1.5}}{\sqrt{LC}}$
we know that $W _{resonance}=\dfrac{1}{\sqrt{LC}} .... (2)$
Clearly from $(1)$ and $(2)$ $W>W _{res}(C)$

The frequency for which $5\mu F$ capacitor has a reactance of $10,000 \Omega$ is 

  1. $(100/\pi )$ cycles per second

  2. $(10/\pi )$ cycles per second

  3. $200$ cycles per second

  4. $5,000$ cycles per second


Correct Option: B
Explanation:

$\ = 5 \times {10^{ - 6}}f$

${x _c} = 10000\Omega $
${x _c} = \dfrac{1}{{{W _C}}}$
$ \Rightarrow w = \dfrac{1}{{c \times {x _c}}} = \dfrac{1}{{5 \times {{10}^{ - 6}} \times 10000}}$
$ = \dfrac{{{{10}^6}}}{{5 \times {{10}^4}}} = \dfrac{{100}}{5} = 20\,\,rad/\sec $
$ \Rightarrow 2\pi f = 20\,\,rad/\sec $
$ \Rightarrow f = \dfrac{{10}}{\pi }\,\,cycles/\sec $
Hence,
option $(B)$ is correct answer.

Two point charges $17.7 \mu c$ and $-17,7 \mu c$ separated by a very small distance, are kept inside a large hollow metallic sphere. Electric flux emnating through the sphere is :

  1. $ 2 \times 10^6$Vm

  2. $- 2 \times 10^6$Vm

  3. Zero

  4. $ 4 \times 10^6$Vm


Correct Option: C
Explanation:

Since net charge is zero

$\therefore$ Net flux is zero.
Hence,
option $C$ is correct answer.

A parallel plate capacitor has an electric field of $105$V /m between the plates .If the charge on one of the capacitor plate is 1$\mu$C,then the magnitude of the force on each capacitor plate is :

    1. 1 N
    1. 05 N
    1. 5 N
    1. 01 N

Correct Option: B
Explanation:

Force of attraction between the plates capacitor 

$F=\cfrac {1}{2} Q({\dfrac{Q}{A\epsilon _0} })$

=$\cfrac{1}{2} QE$

$\dfrac{1}{2}\times 10^{-6}\times 10^5$

$\dfrac{1}{20}$

$=0.05 N$

In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates.The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of $1.6  \times 10^{-19} C$ the charge of the electron. For this, he won the Nobel prize. Extra electrons on this particular oil drop (given the presently known charge of the electron) are :

  1. $4$

  2. $3$

  3. $5$

  4. $8$


Correct Option: A
Explanation:

Given that electrostatic force is just balancing gravitational force.
$qE = mg$   

$q \times 1.68 \times 10^5 = 1.08 \times 10^{-14} g$
Charge of drop, $q = 6.40 \times 10^{-19} C$
Charge of an electron $=1.6\times 10^{-19}$
Let the no. of electrons on the drop is $n$.
Then $ne=6.40 \times 10^{-19} C$
$n=\dfrac{6.40 \times 10^{-19} }{1.6\times 10^{-19}}=4$

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