Physics

Capacitors and Capacitance

189 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 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
C Correct answer
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.

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

When a metal plate is introduced between the two plates of a charged capacitor and insulated from them, then

  1. the metal plate divides the capacitor into two capacitors connected in parallel to each other

  2. the metal plate divides the capacitor into two capacitors connected in series with each other

  3. the metal plate is equivalent to a dielectric of zero dielectric constant

  4. the metal plate is equivalent to a dielectric of infinite dielectric constant
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

The capacitor is divided into two capacitors joined in series. So (b) is correct. Also as K for a metal is $\infty$, therefore (d) is also correct.

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

Dielectric constant, property of an electrical insulating material (a dielectric) equal to

  1. the ratio of the capacitance of a capacitor filled with the given material to the capacitance of an identical capacitor in a vacuum without the dielectric material.

  2. the ratio of the capacitance of a capacitor filled with the given material to the capacitance of an identical capacitor in a vacuum with the dielectric material.

  3. the capacitance of a capacitor filled with the given material

  4. none of the above

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The capacitance of a parallel plate capacitor filled with the given material of dielectric constant $K$ is given by:

              $C=\dfrac{K\varepsilon _{0}A}{d}$ .....................eq1
And the capacitance of the same capacitor without the given material is given by:
              $C'=\dfrac{\varepsilon _{0}A}{d}$  .....................eq2
Dividing eq1 by eq2,
              $C/C'=K$

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

 The capacitance of a capacitor

  1. filled with a dielectric is lesser than it would be in a vacuum.

  2. filled with a dielectric is greater than it would be in a vacuum.

  3. filled with a dielectric is same as it would be in a vacuum.

  4. none of the above

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The capacitance of a parallel plate capacitor filled with dielectric of dielectric constant $K$ is,

           $C=\dfrac{K\varepsilon _{o}A}{d}$  ..................eq1
and the capacitance of a parallel plate capacitor without dielectric in a vacuum is,
           $C'=\dfrac{\varepsilon _{o}A}{d}$  ...................eq2
Dividing eq1 by eq2 , we get
            $C=KC'$
as $K>1$ , therefore $C>C'$ therefore the capacitance of a capacitor filled with a dielectric is greater than it would be in a vacuum.

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

The voltage can be increased, but electric breakdown will occur if the electric field inside the capacitor becomes too large. The capacity can be increased by 

  1. expanding the electrode areas

  2. reducing the gap between the electrodes

  3. expanding the gap between the electrodes

  4. Both A and B
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The capacitance of an air filled parallel plate capacitor is given by:

            $C=\dfrac{\varepsilon _{0}A}{d}$  ..........................eq1
where $A=$ area of each plate (electrode),
            $d=$ distance between plates (electrodes)
from eq1 ,
           $C\propto A$
It is clear that capacity C can be increased by expanding the electrodes area,
and     $C\propto 1/d$ ,
hence, capacity C can be increased by reducing the gap between the electrodes.

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

A dielectric slab of thickness $6 cm$ is placed between the plates of a parallel plate capacitor. If the distance between plates is reduced by $4 cm$, the capacity of the capacitor remains the same. Find the dielectric constant of the medium.

  1. $2$
  2. $4$
  3. $6$
  4. $3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
 for a parallel plate capacitor 
             C = ε K   A / d
It is possible that we have air/vacuum and then a medium in between the parallel plates.  Initially there is a medium of dielectric constant K and of thickness d.

    Let  K1 be the dielectric constant of the slab.  
     let d1 be the thickness of slab = 6cm.

Formula for capacity with multiple media in between the plates is 
             C =  ε A / [ d1/K1 + d2 / K ]
          total gap =  d1 + d2 = d + 4 cm          =>  d2 = d + 4 cm - 6 cm = d - 2
 
         d / K =  d1/ K1 + d2 / K 
         d/ K = 6 / K1  + (d-2)/K

          d - (d-2)  = 6 K / K1 
                K1 = 3 K
  If there was air originally in between the plates,  K = 1,  then answer is 3.
Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

A parallel plate condenser is charged to $100$ volt. In this context which one of the following statements is true?

  1. The two plates attract each other

  2. There is to force between the plates

  3. The two plates of the condenser repel each other

  4. The force between the plates can be attractive or repulsive depending upon the nature of the dielectric material between the plates
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics electric fields introduction to electrostatic force electric force charging and discharging

A capacitor is a perfect insulator for:

  1. constant direct current

  2. alternating current

  3. direct as well as alternating current

  4. variable direct current.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A capacitor consists of two conductors separated by an insulator. In a DC circuit, it blocks current once charged. In an AC circuit, it allows current to flow due to the changing electric field, but it is still physically an insulator between the plates.

Multiple choice physics energy transformations and energy transfers forms of energy and energy conservation energy for everything forms of energy

A charged parallel plate capacitor of distance ($d$) has ${U} _{0}$ energy. A slab of dielectric constant ($K$) and thickness ($d$) is then introduced between the plates of the capacitor. The new energy of the system is given by:

  1. $K{U} _{0}$
  2. ${K}^{2}{U} _{0}$
  3. $\cfrac{{U} _{0}}{K}$
  4. $\cfrac{{U} _{0}}{{K}^{2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The capacitance of two capacitors are $C _1=(5 \pm 0.1)\mu F$ and $C _2=(10 \pm 0.1)\mu F$, If they are connected in series then the percentage error is 

  1. $3.33 $%
  2. $4.03 $%
  3. $3.0 $%
  4. $4.33 $%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two capacitors are in series, the equivalent capacitance is $C=\dfrac{C _1C _2}{C _1+C _2}$
Thus, $\dfrac{\Delta C}{C}=\dfrac{\Delta C _1}{C _1}+\dfrac{\Delta C _2}{C _2}+\dfrac{\Delta C _1+\Delta C _2}{C _1+C _2}$

The % error, $\dfrac{\Delta C}{C}\times 100=\left(\dfrac{\Delta C _1}{C _1}+\dfrac{\Delta C _2}{C _2}+\dfrac{\Delta C _1+\Delta C _2}{C _1+C _2}\right)\times 100$
                                         $=(\dfrac{0.1}{5}+ \dfrac{0.1}{10}+\dfrac{0.1+0.1}{5+10})\times 100=4.33$%

Multiple choice

A circuit consists of a resistor of resistance (R), an inductor of inductance (L), and a capacitor of capacitance (C). The charge on the capacitor is given by the function (q(t) = Q_0\cos\omega t), where (Q_0) is the initial charge on the capacitor and (\omega) is the angular frequency of the circuit. What is the equation of motion for the charge on the capacitor?

  1. \(LC\frac{d^2q}{dt^2} + RC\frac{dq}{dt} + \frac{1}{C}q = 0\)
  2. \(LC\frac{d^2q}{dt^2} - RC\frac{dq}{dt} + \frac{1}{C}q = 0\)
  3. \(LC\frac{d^2q}{dt^2} + RC\frac{dq}{dt} - \frac{1}{C}q = 0\)
  4. \(LC\frac{d^2q}{dt^2} - RC\frac{dq}{dt} - \frac{1}{C}q = 0\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of motion for the charge on the capacitor in an LRC circuit is given by (LC\frac{d^2q}{dt^2} + RC\frac{dq}{dt} + \frac{1}{C}q = 0). This equation is a second-order linear differential equation.