Physics

Capacitors and Capacitance

186 Questions

Capacitors and capacitance are crucial topics in physics, covering the storage of electric charge and energy. These concepts explore series and parallel combinations, dielectric materials, and capacitive reactance. Questions frequently appear in various competitive engineering and medical entrance exams.

Series and parallel capacitorsParallel plate capacitorsCapacitive reactance formulasDielectrics and permittivitySpherical capacitors

Capacitors and Capacitance Questions

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

two similar capacitor are connected to potential v in  parallel order by separating them and joining them in series

  1. the potential on fees plated will be doubled

  2. the charge on the face free plates will increase

  3. the plates in contact will lose their charge

  4. more energy will be stored in the system.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When two capacitors charged to V are connected in parallel, they have charge Q = CV each. If they are disconnected and reconnected in series, the total potential across the combination is V + V = 2V. Thus, the potential on the free plates is doubled.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is 'C' then the resultant capacitance is :

  1. (n+1)C

  2. (n-1)C

  3. nC

  4. C
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here n plates will make $(n-1)$ capacitors with capacitance $C$ and they are in parallel combination.
Thus, the resultant capacitance $=(n-1)C$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Two capacitors of capacitance $C _1$ and $C _2$ are connected in parallel across a battery. If $Q _1$ and $Q _2$ respectively be the charges on the capacitors, then $\dfrac {Q _1}{Q _2}$ will be equal to :

  1. $\dfrac {C _2}{C _1}$
  2. $\dfrac {C _1}{C _2}$
  3. $\dfrac {C _1^2}{C _2^2}$
  4. $\dfrac {C _2^2}{C _1^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In parallel combination the both capacitors have same potential , V (say).
So, $Q _1=C _1V$ and $Q _2=C _2V$
$\therefore \dfrac{Q _1}{Q _2}=\dfrac{C _1V}{C _2V}=\dfrac{C _1}{C _2}$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

These questions consist of two statements, each printed as assertion and reason. While answering these question you are required to choose any one of the following five responses.

If three capacitors of capacitances $\displaystyle { C } _{ 1 }<{ C } _{ 2 }<{ C } _{ 3 }$ are connected in parallel then their equivalent capacitance $\displaystyle $.
Reason: $\displaystyle \frac { 1 }{ { C } _{ p } } =\frac { 1 }{ { C } _{ 1 } } +\frac { 1 }{ { C } _{ 2 } } +\frac { 1 }{ { C } _{ 3 } } $

  1. If both assertion and reason are true but the reason is the correct explanation

    of assertion.

  2. If both assertion and reason are true but the reason is not the correct explanation

    of assertion.

  3. If assertion is true but reason is false.

  4. If both the assertion and reason are false.

  5. If reason is true but assertion is false.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If three capacitors are joined in parallel then their equivalent capacitor will be less than the least value of capacitor so

$C _p > C _s$
$\dfrac{1}{C _p} = \dfrac{1}{C _1}+\dfrac{1}{C _2}+\dfrac{1}{C _3}$ is false.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Calculate the ratio of the equivalent capacitance of the circuit when two identical capacitors are in series to that when they are in parallel?

  1. $\dfrac{1}{4}$
  2. $\dfrac{1}{2}$
  3. $1$
  4. $2$
  5. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the capacitance of each capacitor be $C$.

Series combination :  Equivalent capacitance      $\dfrac{1}{C _{eq}}=\dfrac{1}{C} +\dfrac{1}{C} $                  $\implies C _{eq} = \dfrac{C}{2}$
Parallel combination :    Equivalent capacitance   $C' _{eq} = C + C = 2C$
$\therefore$   $\dfrac{C _{eq}}{C' _{eq}}  = \dfrac{C/2}{2C}  =\dfrac{1}{4}$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Complete the following statements with an appropriate word /term be filled in the blank space(s).


The equivalent capacitance C for the parallel combination of three capacitance $C _1,C _2$ and $C _3$ is given by ${C} =$..............

  1. $C _{1}+ C _{2}+ C _{3}$
  2. $\dfrac{1}{C _{1}+ C _{2}+ C _{3}}$
  3. $\left ( \cfrac{1}{\cfrac{1}{C _{1}}+\cfrac{1}{C _{2}}+\cfrac{1}{C _{3}}}\right )$
  4. $\left ( \cfrac{1}{C _{1}}+\cfrac{1}{C _{2}}+\cfrac{1}{C _{3}}\right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In parallel, the net capacitance is equal to the sum of individual capacitances. In this case, $C = Q/v = Q / (v _1 +v _2+v _3) = Q/v _1 + Q/v _2+ Q/v _3 = C _1 + C _2+ C _3$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A $1\mu F$ capacitor is charged to 200 V and then connected in parallel (+ve to +ve) with a $4\mu F$ capacitor charged to 100 V. The resultant potential difference is :

  1. 120 V

  2. 60 V

  3. 180 V

  4. 150 V
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total charge of the system $=C _1V _1+C _2V _2 = 600\mu C$
Since the capacitors are connected in parallel , the potential across them should be the same. Let the charge across $1\mu F$ capacitor be $q\mu C$.
$ q/1 = (600-q)/4 \Rightarrow 5q=600 \Rightarrow q =120 $
Potential across the capacitors $=Q/C = 120V$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Three capacitance of capacity $10 \mu F , 5 \mu F $ are connected in parallel. The total capacity will be :

  1. $10 \mu F $
  2. $ 5 \mu F $
  3. $ 20 \mu F $

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equivalent capacitance if they are connected in parallel is given by:

${C _{eq}} = {C _1} + {C _2} + {C _3}$

$= 10 + 5 + 5$

$= 20\;{\rm{\mu F}}$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A parallel plate condenser has two circular metal plates of radius 15 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of $10^12V/ms.$ what is the displacement current?

  1. 0.07$A$
  2. 1.39$A$
  3. 13.9$\mathrm { A }$
  4. 139$\mathrm { A }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} Id={ \in _{ 0 } }\dfrac { { d\phi  } }{ { dt } } ={ \in _{ 0 } }\dfrac { { d\left( { EA } \right)  } }{ { dt } } ={ \in _{ 0 } }A\dfrac { { EA } }{ { dt } } =\in \dfrac { { \pi { r^{ 2 } }dE } }{ { dt } }  \ =8.85\times { 10^{ -12 } }\times 3.14\times 25\times { 10^{ -4 } }\times { 10^{ 12 } } \ =0.07A \end{array}$

Hence,
option $(A)$ is correct answer.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A capacitor is charged by a cell of emf $E$ and the charging battery is then removed. If an identical capacitor is now inserted in the circuit in parallel with the previous capacitor, the potential difference across the new capacitor is :

  1. $2E$
  2. $E$
  3. $E/2$

  4. zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As the battery is disconnected so total is constant. i.e $Q _t=CE$
When a identical capacitor is add in parallel so the total capacitance is $C _t=C+C=2C$.
Now the common potential $\displaystyle =\frac{total \  charge }{total \  capacity}=\frac{CE}{2C}=\frac{E}{2}$

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

The dielectric constant $k$ of a medium can be defined as $k=M/N$, where $N$ is the capacity of a parallel plate capacitor. When the space between plates is filled with air, $M$ is ?

  1. the charge of the capacitor

  2. the capacity of a parallel plate capacitor with a dielectric between the plates

  3. the dielectric intensity in the space between the plates

  4. the P.D. across the plates of the condenser, with a dielectric between them
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The dielectric constant k of a medium is defined as the ratio of the capacitance of a capacitor filled with that medium (M) to the capacitance of the same capacitor with a vacuum or air (N). Therefore, M represents the capacity with the dielectric between the plates.

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

The capacitance of an air filled parallel plate capacitor is $8 pF$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $80 \times 10^{-12}$ farads. 
The dielectric constant of wax is

  1. $5.0$
  2. $10.0$
  3. $8.0$
  4. $4.2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

C_air = epsilon_0 * A / d = 8 pF. C_wax = K * epsilon_0 * A / (d/2) = 2 * K * C_air = 80 pF. Thus 2 * K * 8 = 80, so K = 5.

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

The induced charge on the dielectric of a capacitor of capacitance $4\mu F$ when charged by a battery of $50$V, is (dielectric constant of the dielectric $=4$):

  1. $100\mu C$
  2. $200\mu C$
  3. $50\mu C$
  4. $150\mu C$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Induced charge Q_i = Q_free * (1 - 1/K). Q_free = C_air * V = 4uF * 50V = 200uC. Q_i = 200 * (1 - 1/4) = 200 * 0.75 = 150uC.